\(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Rightarrow7-\left(2x+4\right)=-\left(2x-3\right)\)

\(\Rightarrow-\left(2x-3\right)=-\left(x+4\right)\)

\(\Rightarrow3-2x=-x-4\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\)

a) 7-(2x+4)=-(x+4)

  7 - 2x -4 = -x - 4

  7 - 4 + 4 = 2x - x

  7  =  x

x = 7

b)  (x-1) - (2x - 1) = 9 - x

    x - 1 - 2x +1 = 9 - x

  x - 2x + x = 9

   0x = 9( vô lí)

không có giá trị của x thỏa mãn yêu cầu 

a) 7-(2x+4)=-(x+4) <=> 7-2x-4=-x-4 <=>  7-4+4=-x+2x <=> 7=x <=> x=7

b) (x-1)-(2x-1)=9-x <=> x-1-2x+1=9-x <=>  x-1-2x+1=9-x <=> x-2x+x=9-1+1 <=> 0x=9 (vô lí) => không có giá trị của x thỏa mãn 

câu 1:

a)x-1=5-x\(\Leftrightarrow\)x+x=5+1\(\Leftrightarrow\)2x=6\(\Leftrightarrow\)x=3

Vậy tập nghiệm của PT (a) là S={3}

b)3+x=2-x\(\Leftrightarrow\)x+x=2-3\(\Leftrightarrow\)2x=-1\(\Leftrightarrow\)x=-0,5

Vậy tập nghiệm của PT (b) là:S={-0,5}

câu 2:

a) 3x+7=2x-3\(\Leftrightarrow\)3x-2x=-3-7\(\Leftrightarrow\)x=-10

Vậy tập nghiệm của PT (a) là:S={-10}

b)4-(x-2)=(3-2x)\(\Leftrightarrow\)4-x+2=3-2x\(\Leftrightarrow\)-x+2x=-4+3-2\(\Leftrightarrow\)x=-3

Vậy tập nghiệm của PT (b) là:S={-3}

Câu 3:

a)\(\dfrac{5x-4}{2}=\dfrac{16x+1}{7}\Leftrightarrow\dfrac{7\left(5x-4\right)}{14}=\dfrac{2\left(16x+1\right)}{14}\)

\(\Leftrightarrow\)35x-28=32x+2\(\Leftrightarrow\)35x-32x=2+28\(\Leftrightarrow\)3x=30\(\Leftrightarrow\)x=10

Vậy tập nghiệm của PT (a) là :S={10}

b)\(\dfrac{12x+5}{3}=\dfrac{2x-7}{4}\Leftrightarrow\dfrac{4\left(12x+5\right)}{12}=\dfrac{3\left(2x-7\right)}{12}\)

\(\Leftrightarrow\)48x+20=6x-21\(\Leftrightarrow\)48x-6x=-20-21\(\Leftrightarrow\)42x=-41\(\Leftrightarrow\)x=\(-\dfrac{41}{42}\)

Vậy tập nghiệm của PT (b) là:S={\(-\dfrac{41}{42}\)}

a, \(5\left|2x-1\right|-3=7\Leftrightarrow5\left|2x-1\right|=10\Leftrightarrow\left|2x-1\right|=2\)

TH1 : \(2x-1=2\Leftrightarrow x=\frac{3}{2}\)

TH2 : \(2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)

b, \(\left(2x+3\right)\left(x-2\right)-x^2+4=0\Leftrightarrow\left(2x+3\right)\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x+3-x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)

c, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)

\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

\(\Leftrightarrow\left(4x+14\right)^2=\left(3x+9\right)^2\)

\(\Leftrightarrow\left(4x+14+3x+9\right)\cdot\left(4x+14-3x-9\right)=0\)

\(\Leftrightarrow\left(7x+23\right)\left(x+5\right)=0\)

hay \(x\in\left\{-\dfrac{23}{7};-5\right\}\)

\(\Leftrightarrow\left(8x+14\right)^2=\left(9x+3\right)^2\)

\(\Leftrightarrow8x+14=9x+3\)

\(\Leftrightarrow x=11\)

(x - 1)(2x² - 10) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x^2-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{1;\sqrt{5}\right\}\)
(2x - 7)² - 6(2x - 7)(x - 3) = 0

\(\Leftrightarrow\left(2x-7\right)\left(2x-7-6x+18\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(11-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\11-4x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\4x=11\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=\frac{11}{4}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\frac{7}{2};\frac{11}{4}\right\}\)
(5x + 3)(x² + 4) = 0

\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\x^2+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=-3\\x^2=-4\left(Loại\right)\end{matrix}\right.\)

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

Vậy phương trình có tập nghiệm là: \(S=\left\{-\frac{3}{5}\right\}\)

a)

\(\left(x-1\right)\cdot\left(2x^2-10\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot2\cdot\left(x^2-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x^2-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\pm\sqrt{5}\end{matrix}\right.\)

b)

\(\left(2x-7\right)^2-6\cdot\left(6x-7\right)\cdot\left(x-3\right)=0\\ \Leftrightarrow\left(2x-7\right)\cdot\left[\left(2x-7\right)-6\cdot\left(x-3\right)\right]=0\\ \Leftrightarrow\left(2x-7\right)\cdot\left(2x-7-6x+18\right)=0\\ \Leftrightarrow\left(2x-7\right)\cdot\left(11-4x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-7=0\\11-4x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=\frac{11}{4}\end{matrix}\right.\)

c)

\(\left(5x+3\right)\cdot\left(x^2+4\right)=0\)

Vì \(\left(x^2+4\right)>0\Rightarrow\left(loại\right)\)

\(\Rightarrow5x+3=0\\ \Rightarrow x=-\frac{3}{5}\)

| 2-4x | = 4x-2

<=> \(\orbr{\begin{cases}\left|2-4x\right|=-2+4x=4x-2\\\left|2-4x\right|=2-4x=4x-2\end{cases}}\)

<=>\(\orbr{\begin{cases}-2+4x=4x-2\\2-4x=4x-2\end{cases}}\)

<=>\(\orbr{\begin{cases}-2+4x-4x+2=0\\2-4x-4x+2=0\end{cases}}\)

<=>\(\orbr{\begin{cases}0=0\\-8x+4=0\end{cases}}\)

<=> x=\(\frac{-4}{-8}=\frac{1}{2}\)

=> \(S=\left\{\frac{1}{2};\infty\right\}\)

2x-7> 3(x-1)

<=>2x-7>3x-3

<=>2x-3x>-3+7

<=>-x>4

<=>x<4

=>S={x/x<4}

1-2x<4(3x-2)

<=>1-2x<12x-8

<=>-2x-12x<-8-1

<=>-14x<-9

<=>x>\(\frac{9}{14}\)

=>S={\(\frac{9}{14}\)}

-3x+2|-4 -x|> 0

<=>\(\orbr{\begin{cases}-3x+2+4+x>0\\-3x+2-4x-x>0\end{cases}}\)

<=>\(\orbr{\begin{cases}-2x+6>0\\-8x+2>0\end{cases}}\)

<=>\(\orbr{\begin{cases}-2x>-6\\-8x>-2\end{cases}}\)

<=>\(\orbr{\begin{cases}x< 3\\x< \frac{1}{4}\end{cases}}\)

=>S={x/x<3;x/x<\(\frac{1}{4}\)}

4x-1|x-2|< 0

<=>\(\orbr{\begin{cases}4x-1-x+2< 0\\4x-1+x-2< 0\end{cases}}\)

<=>\(\orbr{\begin{cases}3x+1< 0\\3x-3< 0\end{cases}}\)

<=>\(\orbr{\begin{cases}3x< -1\\3x< 3\end{cases}}\)

<=>\(\orbr{\begin{cases}x< \frac{-1}{3}\\x< 1\end{cases}}\)

=>S={x/x<\(\frac{-1}{3}\);x/x<1}

\(\left(3x+1\right)\left(x-3\right)=\left(3x+1\right)\left(2x-5\right)\)

\(\Leftrightarrow\left(3x+1\right)\left(x-3\right)-\left(3x+1\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x-3-2x+5\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(2-x\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}3x+1=0\\2-x=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left[\begin{matrix}x=-\frac{1}{3}\\x=2\end{matrix}\right.\)

Vậy tập nghiệm của pt là \(S=\left\{-\frac{1}{3};2\right\}\)

Có : \(\left(3x+1\right)\left(x-3\right)=\left(3x+1\right)\left(2x-5\right)\)

\(\Leftrightarrow\) \(\left(3x+1\right)\left(x-3\right)-\left(3x+1\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\) \(\left(3x+1\right)\left(x-3-2x+5\right)=0\)

\(\Leftrightarrow\) \(\left(3x+1\right)\left(-x+2\right)=0\)

\(\Leftrightarrow\) \(\left[\begin{matrix}3x+1=0\\-x+2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}3x=-1\\-x=-2\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}x=\frac{-1}{3}\\x=2\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{-1}{3};2\right\}\)

\(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

\(\Leftrightarrow\frac{24\left(x-3\right)}{24}-\frac{4\left(x-3\right)\left(2x-5\right)}{24}=-\frac{6\left(x-3\right)\left(x-3\right)}{24}\)

\(\Leftrightarrow24\left(x-3\right)-4\left(x-3\right)\left(2x-5\right)+6\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)\left[24-4\left(2x-5\right)+6\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(24-8x+20+6x-18\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(26-2x\right)=0\)

\(\Leftrightarrow2\left(x-3\right)\left(13-x\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x-3=0\\13-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=3\\x=13\end{matrix}\right.\)

Vậy tập nghiệm của pt là \(S=\left\{3;13\right\}\)