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a) em thử nhá! Ko chắc đâu
ĐK: \(x\ge-1\)
PT \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=4\Leftrightarrow\sqrt{x+1}=4\Leftrightarrow x=15\) (TMĐK)
b) Do VP >0 nên VT > 0 do đó \(3x-2>x+7\Leftrightarrow x>\frac{9}{2}\)
PT \(\Leftrightarrow\sqrt{3x-2}-5-\left(\sqrt{x+7}-4\right)=0\)
Nhân liên hợp zô: \(\frac{3\left(x-9\right)}{\sqrt{3x-2}+5}-\frac{x-9}{\sqrt{x+7}+4}=0\)
\(\Leftrightarrow\left(x-9\right)\left(\frac{3}{\sqrt{3x-2}+5}-\frac{1}{\sqrt{x+7}+4}\right)=0\)
\(\Leftrightarrow x=9\) (em không biết xử lí cái ngoặc to thế nào đâu..)
\(a,\sqrt{x+1}=\sqrt{2-x}\)
\(\Rightarrow x+1=2-x\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
a) \(ĐKXĐ:-1\le x\le2\)
Bình phương 2 vế ta có:
\(x+1=2-x\)\(\Leftrightarrow2x=1\)\(\Leftrightarrow x=\frac{1}{2}\)( đpcm )
Vậy \(x=\frac{1}{2}\)
b) \(ĐKXĐ:x\ge1\)
\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}+\sqrt{x-1}=16\)
\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
\(\Leftrightarrow2\sqrt{x-1}=16\)\(\Leftrightarrow\sqrt{x-1}=8\)
\(\Leftrightarrow x-1=64\)\(\Leftrightarrow x=65\)( thỏa mãn ĐKXĐ )
Vậy \(x=65\)
c) \(ĐKXĐ:x\ge1\)
\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
\(\Leftrightarrow\sqrt{16\left(x-1\right)}-\sqrt{9\left(x-1\right)}+\sqrt{4\left(x-1\right)}+\sqrt{x-1}=8\)
\(\Leftrightarrow4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8\)
\(\Leftrightarrow4\sqrt{x-1}=8\)\(\Leftrightarrow\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)\(\Leftrightarrow x=5\)( thỏa mãn ĐKXĐ )
Vậy \(x=5\)
Lời giải:
a)
ĐK: \(\forall x\in\mathbb{R}\)
Ta có: \(\sqrt{3x^2}-\sqrt{12}=0\)
\(\Rightarrow \sqrt{3x^2}=\sqrt{12}\)
\(\Rightarrow 3x^2=12\Rightarrow x^2=4\Rightarrow x=\pm 2\) (đều thỏa mãn)
b) ĐK: \(\forall x\in\mathbb{R}\)
\(\sqrt{(x-3)^2}=9\)
\(\Leftrightarrow |x-3|=9\Rightarrow \left[\begin{matrix} x-3=9\\ x-3=-9\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=12\\ x=-6\end{matrix}\right.\)
c) ĐK: $x\in\mathbb{R}$
\(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow \sqrt{(2x)^2+2.2x+1}=6\)
\(\Leftrightarrow \sqrt{(2x+1)^2}=6\)
\(\Leftrightarrow |2x+1|=6\)
\(\Rightarrow \left[\begin{matrix} 2x+1=6\\ 2x+1=-6\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{5}{2}\\ x=-\frac{7}{2}\end{matrix}\right.\)
d) ĐK: \(x\geq 1\)
\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
\(\Leftrightarrow \sqrt{16(x-1)}-\sqrt{9(x-1)}+\sqrt{4(x-1)}+\sqrt{x-1}=8\)
\(\Leftrightarrow 4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8\)
\(\Leftrightarrow 4\sqrt{x-1}=8\Rightarrow \sqrt{x-1}=2\)
\(\Rightarrow x=2^2+1=5\) (thỏa mãn)
e)
ĐK: \(-4\leq x\leq \frac{1}{2}\)
\(\sqrt{1-x}+\sqrt{1-2x}=\sqrt{x+4}\)
\(\Leftrightarrow \sqrt{1-x}-1+\sqrt{1-2x}-1=\sqrt{x+4}-2\)
\(\Leftrightarrow \frac{(1-x)-1}{\sqrt{1-x}+1}+\frac{(1-2x)-1}{\sqrt{1-2x}+1}=\frac{(x+4)-2^2}{\sqrt{x+4}+2}\)
\(\Leftrightarrow \frac{-x}{\sqrt{1-x}+1}+\frac{-2x}{\sqrt{1-2x}+1}=\frac{x}{\sqrt{x+4}+2}\)
\(\Leftrightarrow x\left(\frac{1}{\sqrt{x+4}+2}+\frac{1}{\sqrt{1-x}+1}+\frac{2}{\sqrt{1-2x}+1}\right)=0\)
Dễ thấy biểu thức trong ngoặc lớn lớn hơn $0$
Do đó: \(x=0\) là nghiệm duy nhất của pt.
a) ĐK: \(x\ge5\)
\(\sqrt{4x-20}+\frac{1}{3}\sqrt{9x-45}-\frac{1}{5}\sqrt{16x-80}=0\)
\(\Leftrightarrow\)\(\sqrt{4\left(x-5\right)}+\frac{1}{3}\sqrt{9\left(x-5\right)}-\frac{1}{5}\sqrt{16\left(x-5\right)}=0\)
\(\Leftrightarrow\)\(2\sqrt{x-5}+\sqrt{x-5}-\frac{4}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(\frac{11}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\) (t/m)
Vậy
b) \(-5x+7\sqrt{x}=-12\)
\(\Leftrightarrow\)\(5x-7\sqrt{x}-12=0\)
\(\Leftrightarrow\)\(\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)
đến đây tự làm
c) d) e) bạn bình phương lên
f) \(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^4-2x^2+1\right)+25}\)
\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2}\)
\(\ge\sqrt{9}+\sqrt{25}=8\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x+1=0\\x^2-1=0\end{cases}}\)\(\Leftrightarrow\)\(x=-1\)
Vậy...
Mình làm một vài câu thôi nhé, các câu còn lại tương tự.
Giải:
a) ??? Đề thiếu
b) \(\sqrt{-3x+4}=12\)
\(\Leftrightarrow-3x+4=144\)
\(\Leftrightarrow-3x=140\)
\(\Leftrightarrow x=\dfrac{-140}{3}\)
Vậy ...
c), d), g), h), i), p), q), v), a') Tương tự b)
w), x) Mình đã làm ở đây:
Câu hỏi của Ami Yên - Toán lớp 9 | Học trực tuyến
z) \(\sqrt{16\left(x+1\right)^2}-\sqrt{9\left(x+1\right)^2}=4\)
\(\Leftrightarrow4\left(x+1\right)-3\left(x+1\right)=4\)
\(\Leftrightarrow x+1=4\)
\(\Leftrightarrow x=3\)
Vậy ...
b') \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}=\sqrt{x+1}\)
\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow4\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy ...
a) \(\frac{3}{4}\sqrt{x}-\sqrt{9x}+5=\frac{1}{4}\sqrt{9x}\)
ĐK : x ≥ 0
⇔ \(\frac{3}{4}\sqrt{x}-\sqrt{3^2x}-\frac{1}{4}\sqrt{3^2x}=-5\)
⇔ \(\frac{3}{4}\sqrt{x}-3\sqrt{x}-\frac{1}{4}\cdot3\sqrt{x}=-5\)
⇔ \(-\frac{9}{4}\sqrt{x}-\frac{3}{4}\sqrt{x}=-5\)
⇔ \(-3\sqrt{x}=-5\)
⇔ \(\sqrt{x}=15\)
⇔ \(x=225\)( tm )
b) \(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)
ĐK : x ≤ 3
⇔ \(\sqrt{3-x}-\sqrt{3^2\left(3-x\right)}+\frac{5}{4}\sqrt{4^2\left(3-x\right)}=6\)
⇔ \(\sqrt{3-x}-3\sqrt{3-x}+\frac{5}{4}\cdot4\sqrt{3-x}=6\)
⇔ \(-2\sqrt{3-x}+5\sqrt{3-x}=6\)
⇔ \(3\sqrt{3-x}=6\)
⇔ \(\sqrt{3-x}=2\)
⇔ \(3-x=4\)
⇔ \(x=-1\)( tm )
c) \(\sqrt{9x^2+12x+4}=4\)
⇔ \(\sqrt{\left(3x+2\right)^2}=4\)
⇔ \(\left|3x+2\right|=4\)
⇔ \(\orbr{\begin{cases}3x+2=4\\3x+2=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-2\end{cases}}\)
d) \(\frac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\frac{x-1}{25}}=\frac{29}{15}\)
ĐK : x ≥ 1
⇔ \(\frac{1}{3}\sqrt{x-1}+2\sqrt{2^2\left(x-1\right)}-12\sqrt{\left(\frac{1}{5}\right)^2\cdot\left(x-1\right)}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+2\cdot2\sqrt{x-1}-12\cdot\frac{1}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+4\sqrt{x-1}-\frac{12}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{29}{15}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\sqrt{x-1}=1\)
⇔ \(x-1=1\)
⇔ \(x=2\)( tm )
a) \(\sqrt{4x}=10\) (ĐKXĐ: 4x>=0 <=> x>=0)
\(\Leftrightarrow4x=100\)
\(\Leftrightarrow x=25\)
\(S=\left\{25\right\}\)
b) \(\sqrt{x^2-2x+1}=8\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}=8\)
\(\Leftrightarrow x-1=8\)
\(\Leftrightarrow x=9\)
\(S=\left\{9\right\}\)
c) \(\sqrt{x^2-6x+9}=\sqrt{1-6x+9x^2}\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=\sqrt{\left(1-3x\right)^2}\)
\(\Leftrightarrow x-3=1-3x\) hoặc \(\Leftrightarrow x-3=-1+3x\)
\(\Leftrightarrow x+3x=1+3\) \(\Leftrightarrow x-3x=-1+3\)
\(\Leftrightarrow4x=4\) \(\Leftrightarrow-2x=2\)
\(\Leftrightarrow x=1\) \(\Leftrightarrow x=-1\)
\(S=\left\{1;-1\right\}\)
d) \(\sqrt{2x-5}=x-2\)
\(\Leftrightarrow2x-5=x^2-4x+4\)
\(\Leftrightarrow-x^2+2x+4x-5-4=0\)
\(\Leftrightarrow-x^2+6x-9=0\)
\(\Leftrightarrow x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
\(S=\left\{3\right\}\)
e) \(\sqrt{x^2-2x+1}=\sqrt{x+1}\)
\(\Leftrightarrow x^2-2x+1=x+1\)
\(\Leftrightarrow x^2-2x-x+1-1=0\)
\(\Leftrightarrow x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
\(S=\left\{0;3\right\}\)
g) \(\sqrt{x^2-9}-\sqrt{x-3}=0\) ( ĐKXĐ: x-3>=0 <=> x>=3)
\(\Leftrightarrow\sqrt{x^2-9}=\sqrt{x-3}\)
\(\Leftrightarrow x^2-9=x-3\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow x+2=0\) hoặc \(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=-2\) \(\Leftrightarrow x=3\)
\(S=\left\{-2;3\right\}\)
h) \(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}=1\)
\(\Leftrightarrow x-2+x-3-1=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\)
\(S=\left\{3\right\}\)
i) \(\sqrt{\frac{2x-3}{x-1}}=2\)
\(\Leftrightarrow\frac{2x-3}{x-1}=4\)
\(\Leftrightarrow4\left(x-1\right)=2x-3\)
\(\Leftrightarrow4x-4-2x+3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
\(S=\left\{\frac{1}{2}\right\}\)
l) \(x+y+12=4\sqrt{x}+6\sqrt{y-1}\)
\(\Leftrightarrow x+y-4\sqrt{x}+12-6\sqrt{y-1}=0\)
\(\Leftrightarrow\left(x-4\sqrt{x}+4\right)+\left(y-1-6\sqrt{y-1}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y-1}-3\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\) hoặc \(\Leftrightarrow\sqrt{y-1}-3=0\)
\(\Leftrightarrow\sqrt{x}=2\) \(\Leftrightarrow\sqrt{y-1}=3\)
\(\Leftrightarrow x=4\) \(\Leftrightarrow y-1=9\)
\(\Leftrightarrow y=10\)
KẾT luận : ..............
Tới đây nhé, nếu mai chưa ai giải thì mình giải hộ cho
CHÚC BẠN HỌC TỐT!
m) \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)
<=> \(\sqrt{\left(x-1\right)-4\sqrt{x-1}+4}+\sqrt{\left(x-1\right)+6\sqrt{x-1}+9}=5\)
<=>\(\sqrt{\left(\sqrt{x-1}+2\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}=5\)
<=>\(\sqrt{x-1}+2+\sqrt{x-1}+3=5\)
<=> \(2\sqrt{x-1}=0\)
<=> \(\sqrt{x-1}=0\) <=>x=1
Vậy \(S=\left\{1\right\}\)
n) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\) (*) ( đk \(x\ge\frac{1}{2}\))
<=> \(\left(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}\right)^2=2\)
<=> \(x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{x^2-2x+1}=2\)
<=> 2x+\(2\sqrt{\left(x-1\right)^2=2}\)
<=> x+\(\left|x-1\right|=2\)(1)
TH1: \(\frac{1}{2}\le x\le1\)
Từ (1) => x+1-x=2
<=> 1=2(vô lý)
TH2: x>1
Từ (1)=> x+x-1=2
<=> 2x=3<=> \(x=\frac{2}{3}\)(tm pt (*))
Vậy \(S=\left\{\frac{2}{3}\right\}\)
p) \(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\) (*) (đk :\(x\ge2\))
Đặt \(\left\{{}\begin{matrix}x-2=a\left(a\ge0\right)\\x+1=b\left(b\ge0\right)\end{matrix}\right.\) =>a+b=2x-1
Có \(\sqrt{a+b}+\sqrt{a}=\sqrt{b}\)
<=> \(\sqrt{a+b}=\sqrt{b}-\sqrt{a}\)
<=> \(a+b=b-2\sqrt{ab}+a\)
<=> 0=\(-2\sqrt{ab}\)
=> \(\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\) => x=2 (vì x=-1 không thỏa mãn pt(*))
Vậy \(S=\left\{2\right\}\)
q) \(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\)(*) (đk : \(7\le x\le9\))
Với a,b\(\ge0\) có: \(\sqrt{a}+\sqrt{b}\le2\sqrt{\frac{a+b}{2}}\)(tự cm nha) .Dấu "=" xảy ra <=> a=b
Áp dụng bđt trên có:
\(\sqrt{x-7}+\sqrt{9-x}\le2\sqrt{\frac{x-7+9-x}{2}}=2\sqrt{\frac{2}{2}}=2\) (1)
Có x2-16x+66=(x2-16x+64)+2=(x-8)2+2 \(\ge2\) với mọi x (2)
Từ (1),(2) .Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x-7=9-x\\x-8=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}2x=16\\x=8\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=8\\x=8\end{matrix}\right.\)<=> x=8( tm pt (*))
Vậy \(S=\left\{8\right\}\)
a) ĐKXĐ : \(x\ge-1\)
\(\sqrt{16x+16}-\sqrt{9x+9}=4\)\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=4\)
\(\Leftrightarrow\sqrt{x+1}=4\Leftrightarrow x+1=16\Leftrightarrow x=15\)
b) ĐKXĐ : \(x\ge\frac{2}{3}\)
\(\sqrt{3x-2}-\sqrt{x+7}=1\Leftrightarrow3x-2+x+7-2\sqrt{3x-2}.\sqrt{x+7}=1\)
\(\Leftrightarrow4x+4-2\sqrt{3x^2+19x-14}=0\)\(\Leftrightarrow2x+2-\sqrt{3x^2+19x-14}=0\)
\(\Leftrightarrow2x+2=\sqrt{3x^2+19x-14}\Leftrightarrow\left(2x+2\right)^2=3x^2+19x-14\)
\(\Leftrightarrow4x^2+8x+4=3x^2+19x-14\Leftrightarrow x^2-11x+18=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=9\\x=2\end{cases}\left(tm\right)}\)
c) câu cuối bình phương tương tự câu b
Ai hack nick mình thì trả lại đi !!!
nick :
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