\(\left(2x+1\right)^2=25\)

\(\Rightarrow2x+1\in\left\{-5;5\right\}\)

\(\Rightarrow2x\in\left\{-6;4\right\}\)

\(\Rightarrow x\in\left\{-3;2\right\}\)

Vậy..

\(\left(x-1\right)^3=-125\)

\(\left(x-1\right)^3=-5^3\)

\(x-1=-5\)

\(x=-4\)

Vậy...

\(7^{x+2}.2.7^{x-1}=345\)

\(7^x.\left(7^2+\dfrac{2}{7}\right)=345\)

\(7x=7\)

\(x=1\)

Vậy...

 a) (2x+1)²=25

 (2x+1)²= (+-5)²

=> 2x+1 = 5 hoặc 2x + 1 = -5

2x = 4 hoặc 2x = -6

x= 2 hoặc x=-3

b) (x-1)³=-125

(x-1)³= (-5)³

=> x-1 = -5 

x= -4

c) 2^x+2-2^x=96

2^x.2^{2 -} 2^x = 96

2^x( 4-1) = 96

2^x = 96 : 3

2^x= 32

2^x = 2⁵

=> x= 5

d) 7^x+2+2.7^x-1=345

7^x-1 . 7³ +  2.7^x-1=345

7^x-1( 7³ +2) = 345

7^x-1 . 345 = 345

7^x-1 =1

=> x-1 = 0

=> x= 1

a: \(\left(2x-3\right)^2=\left|3-2x\right|\)

=>\(\left\{{}\begin{matrix}\left|2x-3\right|>=0\\\left(2x-3\right)^2=\left(2x-3\right)\end{matrix}\right.\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)=0\)

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

=>\(\left(2x-3\right)\left(2x-4\right)=0\)

=>\(\left[{}\begin{matrix}2x-3=0\\2x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)

b: \(\left(x-1\right)^2+\left(2x-1\right)^2=0\)

=>\(x^2-2x+1+4x^2-4x+1=0\)

=>\(5x^2-6x+2=0\)

\(\Delta=\left(-6\right)^2-4\cdot5\cdot2=36-20\cdot2=-4< 0\)

=>Phương trình vô nghiệm

c: ĐKXĐ: x>=0

\(x-2\sqrt{x}=0\)

=>\(\sqrt{x}\cdot\sqrt{x}-2\cdot\sqrt{x}=0\)

=>\(\sqrt{x}\left(\sqrt{x}-2\right)=0\)

=>\(\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

d: \(\left(x-1\right)^2+\dfrac{1}{7}=0\)

mà \(\left(x-1\right)^2+\dfrac{1}{7}>=\dfrac{1}{7}>0\forall x\)

nên \(x\in\varnothing\)

a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)

=>\(\left|2x+1\right|=\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

c: (2x-3)²=36

=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

d: \(7^{x+2}+2\cdot7^x=357\)

=>\(7^x\cdot49+7^x\cdot2=357\)

=>\(7^x=7\)

=>x=1

a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

\(---\)

b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)

\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)

\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

\(---\)

c) \(\left(2x-3\right)^2=36\)

\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(---\)

d) \(7^{x+2}+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)

\(\Rightarrow7^x\cdot\left(49+2\right)=357\)

\(\Rightarrow7^x\cdot51=357\)

\(\Rightarrow7^x=357:51\)

\(\Rightarrow7^x=7\)

\(\Rightarrow x=1\)

c) 5^x+5^x+2=650

=> 5^x+5^x5² = 650

=> 5^x ( 1+ 25 ) = 650 

=> 5^x          = 650 / 26

=> 5^x = 25 = 5²

=> x = 2

6) \(\frac{1}{2}.2x+2^{x+2}=2^8+5\)

\(\Rightarrow x+2^{x+2}=2^8+2^5=288\)

- Nếu x < 6 thì x + 2^x+2 < 262

- Nếu x > 6 thì x + 2^x+2 > 519

Vậy không có giá trị nào của x thỏa mãn

b)  \(7^{x+2}+2.7^{x-1}=7^{x-1}.\left(7^3+2\right)=7^{x-1}.345=345\)

\(\Rightarrow7^{x-1}=1\Rightarrow x-1=0\Rightarrow x=1\)

Vậy x = 1 thỏa mãn

a) \(\left(2x+1\right)^2=25\)

=> \(2x+1=5\)         và       \(2x+1=-5\)

=>  \(2x=5-1=4\) và     \(2x=-5-1=-6\)

=>  \(x=4:2=2\) và   \(x=-6:2=-3\)

b) \(\left(x-1\right)^3=-125\)

=> \(x-1=-5\Rightarrow x=-5+1=-4\)

c)  \(2^{x+2}-2^x=96\)

=> \(2^x\cdot2^2-2^x\cdot1=96\)

=> \(2^x\left(2^2-1\right)=96\)

=> \(2^x\cdot3=96\Rightarrow2^x=96:2=32\)

=> \(x=5\)

d) \(7^{x+2}+2\cdot7^{x-1}=345\)

=> \(7^x\cdot7^2+2\cdot7^x:7=345\)

=> \(7^x\cdot7^2+2\cdot7^x\cdot\frac{1}{7}=345\)

=> \(7^x\cdot\left(7^2+2\cdot\frac{1}{7}\right)=345\)

=> \(7^x\cdot\frac{345}{7}=345\)

=> \(7^x=345:\frac{345}{7}=7\)

=> \(x=1\)

\(\left(2x+1\right)^2=25\)

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

\(TH1:\left(2x+1\right)^2=5^2\)

\(2x+1=5\)

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

\(x=4\)

\(TH2:\left(2x+1\right)^2=\left(-5\right)^2\)

\(2x+1=-5\)

\(x=\left[\left(-5\right)-1\right]:2\)

\(x=-3\)

Vậy x=2 hoặc x= -3

Lời giải:

a.

$|2x-5|=12-3x$

Nếu $x\geq \frac{5}{2}$ thì $2x-5=12-3x$

$\Leftrightarrow x=3,4$ (thỏa mãn)

Nếu $x< \frac{5}{2}$ thì: $5-2x=12-3x$

$\Leftrightarrow x=7$ (loại)

Vậy......

b.

$4x=|x+1|+|x+2|+|x+3|\geq 0$

$\Rightarrow x\geq 0$

Do đó: $|x+1|+|x+2|+|x+3|=(x+1)+(x+2)+(x+3)=3x+6$

Vậy: $3x+6=4x$

$\Leftrightarrow x=6$ (thỏa mãn)

c.

$|x^2+|x+2||=x^2+3$

$\Leftrightarrow x^2+|x+2|=x^2+3$
$\Leftrightarrow |x+2|=3$

$\Leftrightarrow x+2=3$ hoặc $x+2=-3$

$\Leftrightarrow x=1$ hoặc $x=-5$

d.

$|x^2-3|=6$

$\Leftrightarrow x^2-3=6$ hoặc $x^2-3=-6$

$\Leftrightarrow x^2=9$ (chọn) hoặc $x^2=-3< 0$ (loại)

$\Leftrightarrow x=\pm 3$

\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)

\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)

\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)

hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)