Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\sqrt{4x+20}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)\(\sqrt{9x-45}\)=4 ; ĐKXĐ : x ≥_+ 5
⇔ \(\sqrt{2^2x+2^2.5}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)\(\sqrt{3^2x-3^2.5}\) =4
⇔ 2\(\sqrt{x+5}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)3\(\sqrt{x-5}\) =4 ⇔ 2\(\sqrt{x+5}\) +\(\sqrt{x-5}\) -\(\sqrt{x-5}\) =4⇔2\(\sqrt{x+5}\)=4(tm)
⇔\(\sqrt{x+5}\)=2⇔x+5=4 ⇔x=-1
Vậy x=-1
b) \(\sqrt{x^2-36}\) - \(\sqrt{x-6}\) =0 ; ĐKXĐ: x≥_+6
⇔ \(\sqrt{\left(x-6\right)\left(x+6\right)}\) - \(\sqrt{x-6}\) =0 ⇔ \(\sqrt{x-6}\).\(\sqrt{x+6}\) - \(\sqrt{x-6}\) =0
⇔ \(\sqrt{x-6}\)(\(\sqrt{x+6}\) -1 )=0 ⇔\([\) \(\begin{matrix}\sqrt{x-6}&=0\\\sqrt{x+6}-1&=0\end{matrix}\) ⇔ \([\) \(\begin{matrix}x-6&=0\\x+6-1&=0\end{matrix}\) ⇔\([\) \(\begin{matrix}x&=6\left(ktm\right)\\x&=-5\left(tm\right)\end{matrix}\)
Vậy x=-5
c) \(\sqrt{4-x^2}\) -x +2 =0 ; ĐKXĐ: -2≤x≤2
⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) -x+2 =0 ⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) -(x-2)=0
⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) =(x-2) ⇔ (2-x)(2+x)=(x-2)2 ⇔ 4-x2 = x2-4x+4 ⇔ -x2-x2+4x=4-4
⇔-2x2+4x=0 ⇔ -2x(x-2)=0 ⇔ \([\) \(\begin{matrix}-2x&=0\\x-2&=0\end{matrix}\) ⇔\([\) \(\begin{matrix}x&=0\left(tm\right)\\x&=2\left(tm\right)\end{matrix}\)
Vậy S=\(\left\{0;2\right\}\)
d) \(\sqrt{\left(2x-3\right)\left(x-1\right)}-\sqrt{x-1}=0\) ; ĐKXĐ: x≥\(\dfrac{3}{2}\);x ≥ 1
⇔\(\sqrt{2x-3}.\sqrt{x-1}-\sqrt{x-1}=0\) ⇔ \(\sqrt{x-1}.\left(\sqrt{2x-3}-1\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{2x-3}-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x-1=0\\2x-3-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Vậy s=\(\left\{1:2\right\}\)
b) Cách làm cũng giống như thế :v
ĐKXĐ: \(x\ge\frac{1}{2}\)
\(PT\Leftrightarrow\left(x-1\right)\left(\frac{4x+6}{\sqrt{2x-1}+1}+\frac{x}{\sqrt{x+3}+2}+x\right)=0\)
\(\Leftrightarrow x=1\) (TMĐK)
a) ĐKXĐ: \(x\ge1\).
\(PT\Leftrightarrow x\left(\sqrt{x-1}-1\right)+\left(2x+1\right)\left(\sqrt{x+2}-2\right)+\left(x^3-4x^2+6x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{x}{\sqrt{x-1}+1}+\frac{2x+1}{\sqrt{x+2}+2}+x^2-2x+2\right)=0\)
\(\Leftrightarrow x=2\left(TMĐK\right)\)
\(x^2-2-2\sqrt{4x-7}=0\)
\(\Leftrightarrow\left(4x-7-2\sqrt{4x-7}+1\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(\sqrt{4x-7}-1\right)^2+\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{4x-7}-1=0\\x-2=0\end{matrix}\right.\)
Tự làm tiếp nhé.
. . .
\(4x^2-5x+1+2\sqrt{x-1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)+2\sqrt{x-1}=0\)
\(\Leftrightarrow\sqrt{x-1}\left[\left(4x-1\right)\sqrt{x-1}+2\right]=0\)
\(\Rightarrow x=1\)
. . .
\(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}=1\)
\(\Leftrightarrow\left|x-2\right|+\left|x-3\right|=1\)
\(VT=\left|x-2\right|+\left|3-x\right|\ge\left|x-2+3-x\right|=1=VP\)
Dấu "=" xảy ra khi \(\left(x-2\right)\left(3-x\right)\ge0\)
Đến đây lập bảng xét dấu
. . .
\(x^2-x+2=2\sqrt{x^2-x+1}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+1}-1\right)^2=0\)
Tự làm tiếp nhé.
\(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)
\(\Leftrightarrow\left(\sqrt{3x+1}-4\right)+\left(1-\sqrt{6-x}\right)+\left(3x^2-14-5\right)=0\)
\(\Leftrightarrow\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{1-6+x}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1\right)\left(x-5\right)=0\)
\(\Rightarrow x=5\)
. . .
\(\sqrt{2x^2-4x+5}-x+4=0\)
\(\Leftrightarrow\sqrt{2x^2-4x+5}=x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-4\ge0\\2x^2-4x+5=x^2-8x+16\end{matrix}\right.\)
Tự làm tiếp nhé.
. . .
\(\sqrt{2x+3}+\sqrt{x-1}=\sqrt{x+6}\)
\(\Leftrightarrow\sqrt{2x+3}=\sqrt{x+6}-\sqrt{x-1}\)
\(\Leftrightarrow2x+3=x+6-2\sqrt{\left(x+6\right)\left(x-1\right)}+x-1\)
\(\Leftrightarrow2\sqrt{x^2+5x-6}=2\)
\(\Leftrightarrow x^2+5x-6=1\)
Tự làm tiếp nhé.
. . .
\(x+y+\dfrac{1}{2}=\sqrt{x}+\sqrt{y}\)
\(\Leftrightarrow\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\left(y-\sqrt{y}+\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\left(\sqrt{y}-\dfrac{1}{2}\right)^2=0\)
Tự làm tiếp nhé.
a) \(\left|3x+1\right|=\left|x+1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=x+1\\3x+1=-x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
c) \(\sqrt{9x^2-12x+4}=\sqrt{x^2}\)
\(\Leftrightarrow\sqrt{\left(3x-2\right)^2}=\sqrt{x^2}\)
\(\Leftrightarrow\left|3x-2\right|=\left|x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=x\\3x-2=-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
d) \(\sqrt{x^2+4x+4}=\sqrt{4x^2-12x+9}\)
\(\Leftrightarrow\sqrt{\left(x+2\right)^2}=\sqrt{\left(2x-3\right)^2}\)
\(\Leftrightarrow\left|x+2\right|=\left|2x-3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-3\\x+2=-2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\left|x^2-1\right|+\left|x+1\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-1\)
f) \(\sqrt{x^2-8x+16}+\left|x+2\right|=0\)
\(\Leftrightarrow\sqrt{\left(x-4\right)^2}+\left|x+2\right|=0\)
\(\Leftrightarrow\left|x-4\right|+\left|x+2\right|=0\)
⇒ vô nghiệm
\(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)
\(\Leftrightarrow\sqrt{x-2}-1+\sqrt{4-x}-1=2x^2-5x-3\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{x-2}+1}+\frac{1}{\sqrt{4-x}+1}+2x+1\right)=0\)
\(\Rightarrow x=3\)
phương trình còn lại mk chưa giải đc nhưng nó vô nghiệm
Em thử câu c nha, sai thì thôi
c) ĐK: \(x\ge-1\).Nhận xét x = 0 là không phải nghiệm, xét x khác 0:
Nhân liên hợp ta được \(\left(x+4\right).\left(\frac{x}{\sqrt{x+1}-1}\right)^2=x^2\)
\(\Leftrightarrow\frac{x+4}{\left(\sqrt{x+1}-1\right)^2}=1\Leftrightarrow x+4=\left(\sqrt{x+1}-1\right)^2\)
\(\Leftrightarrow x+4=x+2-2\sqrt{x+1}\) (rút gọn vế phải)
\(\Leftrightarrow\sqrt{x+1}=-1\left(\text{vô lí}\right)\)
Vậy pt vô nghiệm
đặt \(\sqrt{2x-x^2}=a\)
phương trình trở thành:
\(\sqrt{1+a}+\sqrt{1-a}=2\left(1-a^2\right)^2\left(1-2a^2\right)\)
đến đây thì khai triển đi
a) đkxđ: \(\begin{cases}\sqrt{x^2-4}\ge0\\\sqrt{x^2}+4x+4\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}\begin{cases}x-2\ge0\\x+2\ge0\end{cases}\\x+2\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}x\ge2\\x\le-2\end{cases}\) \(\Leftrightarrow-2\ge x\ge2\)
\(\sqrt{x^2-4}+\sqrt{x^2+4x+4}=0\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}+\sqrt{\left(x+2\right)^2}=0\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}=x+2\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x+2\right)^2\)
\(\Leftrightarrow\left(x+2\right)\left(x-2-x+2\right)=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
S={-2}
b) đkxđ: \(\begin{cases}\sqrt{1-x^2}\ge0\\\sqrt{x+1}\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}1-x^2\ge0\\x+1\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}x^2\le1\\x\ge-1\end{cases}\) \(\Leftrightarrow\begin{cases}\begin{cases}x\le1\\x\ge-1\end{cases}\\x\ge-1\end{cases}\) \(\Leftrightarrow-1\le x\le1\)
\(\sqrt{1-x^2}+\sqrt{x+1}=0\)
\(\Leftrightarrow\sqrt{1-x^2}=-\sqrt{x+1}\)
\(\Leftrightarrow1-x^2=x+1\)
\(\Leftrightarrow-x-x^2=0\)
\(\Leftrightarrow-x\left(1+x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}-x=0\\1+x=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\left(N\right)\\x=-1\left(N\right)\end{array}\right.\)
S={-1;0}
\(a,PT\Leftrightarrow\sqrt{x-1-2\sqrt{x-1}+1}=3\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}=3\)
\(\Leftrightarrow\sqrt{x-1}=4\Leftrightarrow x-1=16\Leftrightarrow x=17\)
Vậy............................................
\(b,PT\Leftrightarrow\sqrt{\left(x^2-1\right)^2}=x-1\)
\(\Leftrightarrow x^2-1=x-1\Leftrightarrow x^2=x\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy...............................................
a, \(x-3\sqrt{x}+2=0\Leftrightarrow x-2\sqrt{x}-\sqrt{x}+2=0\)đk : x >= 0
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\Leftrightarrow x=1;x=4\)
b, \(\sqrt{x^2-1}-\sqrt{x+1}=0\Leftrightarrow\sqrt{\left(x-1\right)\left(x+1\right)}-\sqrt{x+1}=0\)đk : \(x\ge1\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x-1}-1\right)=0\)
TH1 : \(x=-1\)( loại )
TH2 : \(\sqrt{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\)
c, \(x^2+4x+4-\sqrt{2x+1}-\left(x-1\right)^2=0\)đk : x>= -1/2
\(\Leftrightarrow\left(x+2\right)^2-\left(x-1\right)^2-\sqrt{2x+1}=0\)
\(\Leftrightarrow3\left(2x+1\right)-\sqrt{2x+1}=0\Leftrightarrow\sqrt{2x+1}\left(3\sqrt{2x+1}-1\right)=0\)
TH1 : \(x=-\frac{1}{2}\)
TH2 : \(\sqrt{2x+1}=\frac{1}{3}\Leftrightarrow2x+1=\frac{1}{9}\Leftrightarrow x=\frac{\frac{1}{9}-1}{2}=\frac{-\frac{8}{9}}{2}=-\frac{4}{9}\)
a) ĐK : x \(\ge0\)
\(x-3\sqrt{x}+2=0\)
<=> \(\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{x}-1=0\\\sqrt{x}-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}}\)(tm)
b) ĐK \(\hept{\begin{cases}x\ge-1\\x\notin\left\{x\in R|-1< x< 0\right\}\end{cases}}\)
\(\sqrt{x^2-1}-\sqrt{x+1}=0\)
<=> \(\sqrt{x-1}\sqrt{x+1}-\sqrt{x+1}=0\)
<=> \(\sqrt{x-1}\left(\sqrt{x+1}-1\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{x+1}=0\\\sqrt{x-1}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)(tm)
c) ĐK : \(x\ge-\frac{1}{2}\)
\(x^2+4x+4-\sqrt{2x+1}-\left(x-1\right)^2=0\)
<=> \(6x+3-\sqrt{2x+1}=0\)
<=> \(\sqrt{2x+1}\left(3\sqrt{2x+1}-1\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{2x+1}=0\\3\sqrt{2x+1}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-\frac{4}{9}\end{cases}}\)(tm)
a) x ϵ { 4;1}
b) x ϵ ϕ
c) x ϵ {\(\dfrac{-1}{2};-1\)}
a/ \(x\in\left\{4;1\right\}\)
b/ x=2
c/\(x\in\left\{-\dfrac{1}{2};-\dfrac{4}{9}\right\}\)
a) S={4;1}
b) S={-1;2}
c) S={\(\dfrac{-1}{2}\);\(\dfrac{-4}{9}\)}
a)\(x\in\left\{4;1\right\}\)
b)\(x\in\left\{-1;2\right\}\)
c)x=\(\dfrac{-4}{9}\)
a) \(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
b) x= 2
c) \(\left[{}\begin{matrix}\dfrac{-1}{2}\\\dfrac{-4}{9}\end{matrix}\right.\)
a) x-3\(\sqrt{x}\) +2 =0\((x\ne0)\) ⇔ x -2\(\sqrt{x}\) -\(\sqrt{x}\) +2=0 ⇔ \(\sqrt{x}\)(\(\sqrt{x}\) -1)-2(\(\sqrt{x}\) - 1)=0
⇔(\(\sqrt{x}\) - 2)(\(\sqrt{x}\) -1)=0 ⇒\([\)\(\begin{matrix}\sqrt{x}-2&=0\\\sqrt{x}-1&=0\end{matrix}\)⇔\([\) \(\begin{matrix}\sqrt{x}&=2\\\sqrt{x}&=1\end{matrix}\)⇒ \([\) \(\begin{matrix}x&=4\left(tm\right)\\x&=1\left(tm\right)\end{matrix}\)
b)
a. \(x-3\sqrt{x}+2=x-2\sqrt{x}-\sqrt{x}+2=\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\)
b. \(\sqrt{x^2-1}-\sqrt{x+1}=\sqrt{\left(x-1\right)\left(x+1\right)}-\sqrt{x+1}=\sqrt{x+1}\left(\sqrt{x-1}-1\right)\)
c. \(x^2+4x+4-\sqrt{2x+1}-\left(x-1\right)^2=x^2+4x+4-\sqrt{2x+1}-x^2+2x-1=6x+3-\sqrt{2x+1}=\sqrt{2x+1}\left(3\sqrt{2x+1}-1\right)\)
a) x1 =1; x2 = 4.
b) x1 =1; x2 =2.
c) x =\(\dfrac{1}{2}\).
a) \(x=1\) hoặc \(x=4\)
b) \(x=4\)
c) \(x=\dfrac{-1}{2}\)
a) S= {1;4}
b) x= 2
c) S={\(\dfrac{-1}{2}\); \(\dfrac{-4}{9}\)}
a)x-3\(\sqrt{x}\)+2=0\(\Leftrightarrow\)(\(\sqrt{x}\)-2)(\(\sqrt{x}\)-1)=0\(\Leftrightarrow\)x=4;x=1
b)\(\sqrt{x^2-1}\)-\(\sqrt{x+1}\)=0\(\Leftrightarrow\)\(\sqrt{\left(x-1\right)\left(x+1\right)}\)-\(\sqrt{x+1}\)=0\(\Leftrightarrow\)\(\sqrt{x+1}\)(x-2)=0\(\Leftrightarrow\)x=-1,x=2
c)x2+4x+4-\(\sqrt{2x+1}\)-(x-1)2=0\(\Leftrightarrow\)6x+3-\(\sqrt{2x+1}\)=0\(\Leftrightarrow\) 3(2x-1)-\(\sqrt{2x-1}\)=0\(\Leftrightarrow\)\(\sqrt{2x+1}\)(3\(\sqrt{2x-1}\))=0\(\Leftrightarrow\)\(\sqrt{2x+1}\)=0,3\(\sqrt{2x-1}\)=0\(\Leftrightarrow\)x=\(\dfrac{-1}{2}\),2x+1=\(\dfrac{1}{9}\)\(\Leftrightarrow\)x=\(\dfrac{-1}{2}\),x=\(\dfrac{-4}{9}\)
A)đkxđ x>=5
X=9(tm)
Vậy pt có 1 nghiệm là S={9}
B)x=6
x=7
Vậy pt có 2 nghiệm S={6;7}
C)đkxđ -2=<x=<2
x=2(tm)
Vậy pt có 1 nghiệm là S={2}
D)đkxđ x>=3/2
x=2(tm)
Vậy pt có nghiệm là S={2}
A)đkxđ x>=0
-x=1(t/m)
-x=4(t/m)
Vậy pt có 2 nghiệm là S={1;4}
B)đkxđ x>=1
x=2(tm)
Vậy pt có 1 nghiệm là S={2}
C)đkxđ x>=-1/2
x=-1/2(tm)
x=-4/9(tm)
a) x - \(3\sqrt{x}\) + 2 = 0 đkxđ: x≥0
<=> ( \(\sqrt{x}\)- 2)(\(\sqrt{x}\) -1)=0
<=> \(\sqrt{x}\) -2=0 hoặc \(\sqrt{x}\) - 1=0
<=> x = 4 (tm) hoặc x = 1 (tm)
vậy S={ 4; 1 }
b) \(\sqrt{x^2-1}-\sqrt{x+1}=0\)
<=> \(\sqrt{x^2-1}=\sqrt{x+1}\) đkxđ: x≥1
<=> \(\left(\sqrt{x^2-1}\right)=\left(\sqrt{x+1}\right)\)
<=> \(x^2-1=x+1\)
<=> \(\left(x+1\right)\left(x-1\right)=x+1\)
<=> (x + 1) (x - 1) - ( x + 1 ) = 0
<=> ( x + 1) ( x -1 - 1 ) = 0\
<=> ( x + 1) ( x - 2 ) = 0
<=> x+1=0 hoặc x-2=0
<=> x=-1 (ktm) hoặc x= 2 (tm)
vâỵ S= { 2 }
c) \(x^2+4x+4-\sqrt{2x+1}-\left(x-1\right)^{^2}=0\) đkxđ: x≥\(\dfrac{-1}{2}\)
<=> \(x^2+4x+4-\sqrt{2x+1}-x^2+2x-1=0\)
<=> \(4x+4-\sqrt{2x+1}+2x-1=0\)
<=> \(6x+3-\sqrt{2x+1}=0\)
<=> \(\sqrt{2x+1}=6x+3\)
<=> \(\left(\sqrt{2x}+1\right)^2=\left(6x+3\right)^2\)
<=> \(2x+1=36x^2+36x+9\)
<=> \(2x+1-36x^2-36x-9=0\)
<=> \(-34x-8-36x^2=0\)
<=>\(18x^2+9x+8x+4=0\)
<=> \(9x.\left(2x+1\right)+4\left(2x+1\right)=0\)
<=> ( 9x+4)(2x+1)=0
<=> 9x+4=0 hoặc 2x+1=0
<=> x=\(\dfrac{-4}{9}\)(tm) hoặc x=\(\dfrac{-1}{2}\)(tm)
vậy S= { \(\dfrac{-4}{9};\dfrac{-1}{2}\) }
a) ĐKXĐ : x≥0
x-√3x+2=0
⇔(√x-1)(√x-2)=0
⇔√x-1=0 or √x-2 =0
⇔√x-1=0
⇔√x=1
⇒x=1(nhận)
⇔√x-2=0
⇔√x=2
⇒x=4(nhận)
→S={1;4}
b)ĐKXĐ=x≥1
√x^2-1-√x+1=0
=√(x-1)(x+1)-√x+1=0
=√x-1.√x+1-√x+1=0
=√x+1(√x-1 -1)=0
⇔√x+1=0 or √x-1 -1
√x+1=0
√x=-1
x=-1(loại)
√x-1 -1=0
√x-1=1
√x=√1+1
x=2(nhận)
⇒Vậy S={2}
c)ĐKXĐ≥-1÷2
x^2+4x+4-√2x+1 -(x-1)^2=0
=(x+2)^2-(x-1)^2- √2x+1=0
=(x+2+x-1)(x+2-x+1)- √2x+1=0
=3(2x+1)- √2x+1=0
=√2x+1(3. √2x+1 -1)=0
== √2x+1=0
=√2x=-1
=x=-1÷2(nhận)
==3. √2x+1 -1 =0
=x=-4÷9(nhận)
---Vậy S={-1÷2;-4÷9}
a) \(x-3\sqrt{x}+2=0\) Điều kiện xác định: \(x\ge0\)
⇔\(x-\sqrt{x}-2\sqrt{x}+2=0\)
⇔\(\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)=0\)
⇔\(\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)
⇔\(\begin{matrix}\sqrt{x}-2=0\\\sqrt{x}-1=0\end{matrix}\Leftrightarrow\begin{matrix}\sqrt{x}=2\\\sqrt{x}=1\end{matrix}\Leftrightarrow\begin{matrix}x=4\\x=1\end{matrix}\)
Vậy pt có nghiệm là (4;1)
b) \(\sqrt{x^2-1}-\sqrt{x-1}=0\)
⇔\(\sqrt{\left(x-1\right)\left(x-1\right)}-\sqrt{x-1}=0\)
⇔\(\sqrt{x-1}\left(\sqrt{x+1}-1\right)=0\)
⇔\(\begin{matrix}\sqrt{x-1}=0\\\sqrt{x+1}-1=0\end{matrix}\Leftrightarrow\begin{matrix}x-1=0\\\sqrt{x+1}=1\end{matrix}\Leftrightarrow\begin{matrix}x=1\\x=0\end{matrix}\)
Vậy pt có nghiệm là (1;0)
c) \(x^2+4x+4-\sqrt{2x+1}-\left(x-1\right)^2=0\) Diều kiện xác định: \(x\ge\dfrac{1}{2}\)
⇔\(\left(x+2\right)^2-\sqrt{2x+1}-\left(x-1\right)^2=0\)
⇔\(\left(x+2-x+1\right)\left(x+2+x-1\right)-\sqrt{2x+1}=0\)
⇔\(3\left(2x+1\right)-\sqrt{2x-1}=0\)
⇔\(\sqrt{2x+1}\left[3\left(\sqrt{2x+1}\right)-1\right]=0\)
⇔\(\begin{matrix}\sqrt{2x+1}=0\\3\left(\sqrt{2x+1}\right)-1=0\end{matrix}\Leftrightarrow\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{-4}{9}\end{matrix}\)
Vậy pt có nghiệm là (\(\dfrac{-1}{2};\dfrac{-4}{9}\))