a/ \(\left|1-2x\right|>7\Leftrightarrow\left[{}\begin{matrix}1-2x=7\\1-2x=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x< -6\\2x< 8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -3\\x< 4\end{matrix}\right.\)

b/ \(\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\) ( vì -5<0)

\(\Leftrightarrow x>3\)

sao ko trả lời câu c

a, -3/4-2x= 4

 2x=-19/4

x=-19/8

(-2/3x-3/5).(-29/6)=2/5

-2/3x-3/5=-12/145

-2/3x=15/29

x=-45/58

c, x/2-3x/5+13/5=-7/5-7/10x

x/2-3x/5+7/10x=-7/5-13/5=-4

5x/10-6x/10+7x/10=-4

6x/10=-4

6x=-40

x=-20/3

a) \(\left(x-1\right)^3=8=2^3\)

\(x-1=2\)

\(x=2+1=3\)

b) \(7^{2x-6}=49=7^2\)

\(2x-6=2\)

\(2x=6+2=8\)

\(x=8:2=4\)

c) \(\left(2x-14\right)^7=128=2^7\)

\(2x-14=2\)

\(2x=14+2=16\)

\(x=16:2=8\)

d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)

\(x=5\)

e) \(3\cdot\left(x+2\right):7\cdot4=120\)

\(x+2=120:3\cdot7:4\)

\(x+2=70\)

\(x=70-2=68\)

Lời giải:

a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$

$\Rightarrow x=3$

b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$

$\Rightarrow 2x=8$

$\Rightarrow x=4$

c. $(2x-14)^7=128=2^7$

$\Rightarrow 2x-14=2$

$\Rightarrow 2x=16$

$\Rightarrow x=18$

d.

$x^4.x^5=5^3.5^6$

$x^9=5^9$

$\Rightarrow x=5$

e. 

$3(x+2):7=120:4=30$

$3(x+2)=30.7=210$

$x+2=210:3=70$

$x=70-2=68$

...

\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)

\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)

1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

cho mình

a,\((x+4)^2-(x+1)(x-1)=16\)

 \(\Rightarrow x^2+8x+16-x^2+1=16\)

\(\Rightarrow 8x=-1\Rightarrow x=-\dfrac{1}{8}\)

b,\((2x-1)^2-(x+3)^2-5(x+7)(x-7)=0\)

\(\Rightarrow 4x^2-4x+1-(x^2+6x+9)-5(x^2-49)=0\)

\(\Rightarrow 4x^2-4x+1-x^2-6x-9-5x^2-245=0\)

\(\Rightarrow -x^2-10x-244=0\)

\(\Rightarrow -(x^2-10x+25)-219=0\)

\(\Rightarrow -(x-5)^2-219=0\)

\(\Rightarrow (x-5)^2+219=0\)

Mà \((x-5)^2+219>0\) suy ra PT vô nghiệm

a) \(25x^2-9=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)

Vậy \(S=\left\{\frac{3}{5};\frac{-3}{5}\right\}\)

b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow\left(x^2+8x+16\right)-\left(x^2-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x+17=16\)

\(\Leftrightarrow8x=-1\)

\(\Leftrightarrow x=-\frac{1}{8}\)

Vậy.........

c)\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left(4x^2-4x+1\right)+\left(x^2+6x+9\right)-5\left(x^2-49\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Leftrightarrow2x=-255\)

\(\Leftrightarrow x=-127,5\)

Vậy.............

có j sai xót mong m.n bỏ qua☺

a) \(25x^2-9=0\)                      

<=> \(\left(5x\right)^2=9\)

<=> \(\left(5x\right)^2=3^2\)

<=> \(5x=3\)

<=> \(x=\frac{3}{5}\)

b) \(\left(x+4\right)^2-\left(x-1\right)\left(x+1\right)=16\)

<=> \(x^2+2.x.4+4^2-\left(x^2-1^2\right)=16\)

<=> \(x^2+8x+16-x^2+1=16\)

<=> \(\left(x^2-x^2\right)+8x+\left(16+1\right)=16\)

<=> \(8x+17=16\)

<=> \(8x=-1\)

<=> \(x=\frac{-1}{8}\)

c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

<=> \(\left(2x\right)^2-2.2x.1+1^2+x^2+2.x.3+3^2-5\left(x^2-7^2\right)=0\)

<=> \(4x^2-4x+1+x^2+6x+9-5x^2+5.7^2=0\)

<=> \(\left(4x^2+x^2-5x^2\right)-\left(4x-6x\right)+\left(1+9+5.7^2\right)=0\)

<=> \(2x+245=0\)

<=> \(2x=-245\)

<=> \(x=\frac{-245}{2}\)

a) \(25x^2-9=0\)

\(\Rightarrow25x^2-3^2=0\)

\(\Rightarrow\left(25x+3\right).\left(25x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}25x+3=0\\25x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}25x=-3\\25x=3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-3}{25}\\x=\frac{3}{25}\end{cases}}\)

b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Rightarrow\left(x+4\right)^2-\left[\left(x+1\right)^2-\left(x-1\right)^2\right]=16\)

\(\Rightarrow\left(x+4\right)^2=16\)

\(\Rightarrow\left(x+4\right)^2=4^2\)

\(\Rightarrow x+4=4\)

\(\Rightarrow x=0\)

c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Rightarrow\left(2x-1\right)+2\left(2x-1\right)\left(x+3\right)+\left(x+3\right)^2-5.\left(x+7\right)^2-\left(x-7\right)^2=0\)

\(\Rightarrow2\left(x+3\right)^3=0\)

\(\Rightarrow\left(x+3\right)^2=0\)

\(\Rightarrow x=3\)

a, |- \(x\) + 2|  - |\(x\) + 7| = 0

           |- \(x\) + 2| = | \(x\) + 7|

           \(\left[{}\begin{matrix}-x+2=x+7\\-x+2=-x-7\end{matrix}\right.\)         

            \(\left[{}\begin{matrix}x=-\dfrac{5}{2}\\2=-7\left(loại\right)\end{matrix}\right.\)

              vậy \(x\) = -\(\dfrac{5}{2}\)

b, |2\(x\) - 1| + |2 + y| ≥ 0

     |2\(x\) - 1| ≥ 0 ∀ \(x\)

     |2 + y| ≥ 0 ∀ y 

  ⇒ |2\(x\) - 1| +|2 + y| ≥ 0 ∀\(x\) ; y