3 x - 2 6 - 5 = 3 - 2 x + 7 4

⇔ 2(3x – 2) – 5.12 = 3[3 – 2(x + 7)]

⇔ 6x – 4 – 60 = 9 – 6(x + 7)

⇔ 6x – 64 = 9 – 6x – 42

⇔ 6x + 6x = 9 – 42 + 64

⇔ 12x = 31 ⇔ x = 31/12

Phương trình có nghiệm x = 31/12

\((3x+\frac{1}{4})-\frac{1}{3}\times(6x+\frac{9}{5})=1\)1

<=> \(3x+\frac{1}{4}-2x-\frac{3}{5}=1\)

<=>\(x-\frac{7}{20}=1\)

<=>\(20x-7=20\)

<=>\(20x=20+7\)

<=>\(20x=27\)

<=> \(x=\frac{27}{20}\)

\(\left(3x+\frac{1}{4}\right)-\frac{1}{3}.\left(6x+\frac{9}{5}\right)=1\)

\(\Leftrightarrow3x+\frac{1}{4}-2x-\frac{3}{5}=1\)

\(\Leftrightarrow x=1+\frac{3}{5}-\frac{1}{4}\)

\(\Leftrightarrow x=\frac{27}{20}\)

1: 5(2-3x)(x-2)=3(1-3x)

=>5(3x-2)(x-2)=3(3x-1)

=>\(5\cdot\left(3x^2-6x-2x+4\right)=9x-3\)

=>\(15x^2-40x+20-9x+3=0\)

=>\(15x^2-49x+23=0\)

\(\Delta=49^2-4\cdot15\cdot23=1021>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{49-\sqrt{1021}}{2\cdot15}=\frac{49-\sqrt{1021}}{30}\\ x=\frac{49+\sqrt{1021}}{2\cdot15}=\frac{49+\sqrt{1021}}{30}\end{array}\right.\)

2: \(4x^2+4x+1=0\)

=>\(\left(2x+1\right)^2=0\)

=>2x+1=0

=>2x=-1

=>\(x=-\frac12\)

3: \(4x^2-9=0\)

=>\(4x^2=9\)

=>\(x^2=\frac94\)

=>\(\left[\begin{array}{l}x=\frac32\\ x=-\frac32\end{array}\right.\)

4: \(5x^2-10x=0\)

=>5x(x-2)=0

=>x(x-2)=0

=>\(\left[\begin{array}{l}x=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)

5: \(x^2-3x=-2\)

=>\(x^2-3x+2=0\)

=>(x-1)(x-2)=0

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)

6: |x-5|-3=0

=>|x-5|=3

=>\(\left[\begin{array}{l}x-5=3\\ x-5=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=2\end{array}\right.\)

AI XEM RỒI NHỚ CHẤM ĐIỂM

Trình bày xấu chưa từng thấy

Thay thế t cho x\(^2\)

Ta có: \(3t^2-12t+9=0\)

Áp dụng phương trình bặc 2:

t=\(\dfrac{-\left(-12\right)\pm\sqrt{\left(-12\right)^2-4.3.9}}{2.3}\)

t=\(\dfrac{12\pm6}{6}\)

=>\(\left\{{}\begin{matrix}t=3\\t=1\end{matrix}\right.\)

Vì thay x\(^2\)=t nên giá trị x=\(\sqrt{t}\)

=>\(\left\{{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\\x=1\\x=-1\end{matrix}\right.\)

Mình mới lớp 8 nên chưa quen giải PT bậc 2, nếu có sai sót mong bạn thông cảm

\(\left\{{}\begin{matrix}2x-2y=-4\\x+2y=-1\end{matrix}\right.\)

⇒ \(3x=-5\)

⇒ \(x=-\dfrac{5}{3}\)

\(a,\left\{{}\begin{matrix}2x-2y=-4\\x+2y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-2y+x+2y=\left(-4\right)+\left(-1\right)\\x+2y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=-5\\x+2y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\-\dfrac{5}{3}+2y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\2y=\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)

\(b,\left\{{}\begin{matrix}3x+5y=11\\2x+5y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y=11\\3x+5y-2x-5y=11-9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3.2+5y=11\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6+5y=11\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5y=5\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)

Không ạ

a: ĐKXĐ: x>=9/4

\(\sqrt{4x-9}=2x-5\)

=>\(\begin{cases}\left(2x-5\right)^2=4x-9\\ 2x-5\ge0\end{cases}\Rightarrow\begin{cases}4x^2-20x+25-4x+9=0\\ x\ge\frac52\end{cases}\)

=>\(\begin{cases}4x^2-24x+34=0\\ x\ge\frac52\end{cases}\Rightarrow\begin{cases}2x^2-12x+17=0\\ x\ge\frac52\end{cases}\)

=>\(\begin{cases}x^2-6x+\frac{17}{2}=0\\ x\ge\frac52\end{cases}\Rightarrow\begin{cases}x^2-6x+9-\frac12=0\\ x\ge\frac52\end{cases}\)

=>\(\begin{cases}\left(x-3\right)^2=\frac12\\ x\ge\frac52\end{cases}\Rightarrow\begin{cases}x-3\in\left\lbrace\frac{\sqrt2}{2};-\frac{\sqrt2}{2}\right\rbrace\\ x\ge\frac52\end{cases}\)

=>\(x=3+\frac{\sqrt2}{2}=\frac{6+\sqrt2}{2}\)

b: ĐKXĐ: \(x^2-7x+10\ge0\)

=>(x-5)(x-2)>=0

=>x>=5 hoặc x<=2

\(\sqrt{x^2-7x+10}=3x-1\)

=>\(\begin{cases}3x-1\ge0\\ \left(3x-1\right)^2=x^2-7x+10\end{cases}\Rightarrow\begin{cases}9x^2-6x+1-x^2+7x-10=0\\ x\ge\frac13\end{cases}\)

=>\(\begin{cases}8x^2+x-9=0\\ x\ge\frac13\end{cases}\Rightarrow\begin{cases}8x^2+9x-8x-9=0\\ x\ge\frac13\end{cases}\)

=>\(\begin{cases}\left(x+1\right)\left(8x-9\right)=0\\ x\ge\frac13\end{cases}\Rightarrow x=\frac98\) (nhận)

d: |3x-1|=x+3

=>\(\begin{cases}x+3\ge0\\ \left(3x-1\right)^2=\left(x+3\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-3\\ \left(3x-1-x-3\right)\left(3x-1+x+3\right)=0\end{cases}\)

=>\(\begin{cases}x\ge-3\\ \left(2x-4\right)\left(4x+2\right\rbrace\end{cases}\Rightarrow x\in\left\lbrace2;-\frac12\right\rbrace\)

e: |x+2|=|6-3x|

=>|3x-6|=|x+2|

=>3x-6=x+2 hoặc 3x-6=-x-2

=>2x=8 hoặc 4x=4

=>x=4 hoặc x=1

\(\sqrt{x^2-6x+9}-4=3x\left(đkxđ:x\ge-\dfrac{4}{3}\right)\\ \Leftrightarrow\sqrt{\left(x-3\right)^2}=3x+4\\ \Leftrightarrow\left|x-3\right|=3x+4\\ \Leftrightarrow\left[{}\begin{matrix}x-3=3x+4\\x-3=-3x-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-3x=4+3\\x+3x=-4+3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-2x=7\\4x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\left(ktm\right)\\x=-\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)

Chị xem lại dòng thứ 4 ạ :>