a) \(\left(2x+1\right).\left(2x-3\right)=7\)

Ta dễ dàng nhìn thấy cho dù x là bất cứ số gì thì phép tính  ( 2x + 1) (2x - 3) vẫn không thể ra 7 (Đây là một điều vô lý)

  Suy ra x vô nghiệm

b) \(x\left(x-7\right)+3.\left(x-7\right)=11\)

\(\Leftrightarrow x\left(x-7\right)^2+3=11\)

 \(\Rightarrow x\left(x-7\right)^2=11-3\)

 \(\Rightarrow x\left(x-7\right)^2=8\)

\(\Rightarrow x=8:\left(x-7\right)^2\)

\(\Rightarrow x=8:\left(x^2-7^2\right)\Leftrightarrow x=8:\left(x^2-49\right)\)

      Bạn tự tính tiếp nhá

a. Đề \(\Leftrightarrow\orbr{\begin{cases}x+11=3-2x\left(x\ge11\right)\\x+11=2x-3\left(x< 11\right)\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x+2x=3-11\left(x\ge11\right)\\2x-x=-3-11\left(x< 11\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=-8\left(x\ge11\right)\\x=-14\left(x< 11\right)\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{8}{3}\left(x\ge11\right)loại\\x=-14\left(x< 11\right)chọn\end{cases}}\)

Vậy \(x=-14\)

Câu b tương tự

Sao lại sai hả bạn?

thì thấy kết quả là khác thì ns sai thôi

`a)-5/6-x=7/12+[-1]/3`

   `-5/6-x=1/4`

   `x=-5/6-1/4`

   `x=-13/12`

`b)(4,5-2x).(-1 4/7)=11/14`

   `(4,5-2x).(-11/7)=11/14`

    `4,5-2x=11/14:(-11/7)=-0,5`

    `2x=4,5-(-0,5)=5`

    `x=5:2=2,5`

`c)(2/11+1/3)x=(1/7-1/8).56`

   `17/33 x=1/56 .56=1`

      `x=33/17`

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam

- Mình có thể giúp bài 1 :)

 5^x +  5^ ( x + 2 ) = 650

5^x +  5^x . 5² = 650

 5^x .( 1 + 25 ) = 650

 5^x . 26 = 650 

5^x = 650 : 26

 5^x = 25

 5^x = 5² 

=> x = 2 

Vậy x = 2 

The Reflection Of Light~ Ủa đề  là 5^x-5^x+2= 650 sao ông lại đổi thành 5^x-5^(x+2) zậy

a) Ta có: \(3\left(x+5\right)-3=2\left(x+1\right)+7\)

\(\Leftrightarrow3x+15-3=2x+2+7\)

\(\Leftrightarrow3x+12=2x+9\)

hay x=-3

b) Ta có: \(15-\left(x+2\right)^2=-1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

c) Ta có: \(75-\left(2x+1\right)^3=11\)

\(\Leftrightarrow\left(2x+1\right)^3=64\)

\(\Leftrightarrow2x+1=4\)

hay \(x=\dfrac{3}{2}\)

a) Ta có: $3\left(x+5\right)-3=2\left(x+1\right)+7$

$\iff 3x+15-3=2x+2+7$

$\iff 3x+12=2x+9$

hay x=-3

b) Ta có: $15-{}^{}=-1$

$\iff {}^{}=16$

⇔[x+2=4x+2=−4⇔[x=2x=−6⇔[x+2=4x+2=−4⇔[x=2x=−6

c) Ta có: $75-{}^{}$

`3(x+5)-3=2(x+1)+7`

`3x+15-3=2x+2+7`

`x=-3`

.

`15-(x+2)^2=-1`

`(x+2)^2=16`

`(x+2)^2=4^2=(-4)^2`

`[(x+2=4),(x+2=-4):}`

`[(x=2),(x=-6):}`

.

`75-(2x+1)^3=11`

`(2x+1)^3=64`

`(2x+1)^2=4^3`

`2x+1=4`

`x=3/2`

$3(x+5)-3=2(x+1)+7$

$3x+15-3=2x+2+7$

$x=-3$

.

$15-{}^{}=-1$

${}^{}=16$

${}^{}={}^{}={}^{}$

$[\begin{array}{c}x+2=4\\ x+2=-4\end{array}$

$[\begin{array}{c}x=2\\ x=-6\end{array}$

a) Ta có: \(3\left(x+5\right)-3=2\left(x+1\right)+7\)

\(\Leftrightarrow3x-2x=2+7-15+3\)

hay x=-3

b) Ta có: \(15-\left(x+2\right)^2=-1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

\(a)\) \(\left(2x+1\right)\left(2x-3\right)=7\)

Có \(4\) trường hợp : 

\(\Leftrightarrow\)\(\hept{\begin{cases}2x+1=1\\2x-3=7\end{cases}\Leftrightarrow\hept{\begin{cases}2x=0\\2x=10\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x=5\end{cases}}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}2x+1=-1\\2x-3=-7\end{cases}\Leftrightarrow\hept{\begin{cases}2x=-2\\2x=-4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\x=-2\end{cases}}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}2x+1=7\\2x-3=1\end{cases}\Leftrightarrow\hept{\begin{cases}2x=6\\2x=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\x=2\end{cases}}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}2x+1=-7\\2x-3=-1\end{cases}\Leftrightarrow\hept{\begin{cases}2x=-8\\2x=2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-4\\x=1\end{cases}}}\)

Vậy không có giá trị nào của x thoả mãn đề bài 

\(b)\) \(x\left(x-7\right)+3\left(x-7\right)=11\)

\(\Leftrightarrow\)\(\left(x+3\right)\left(x-7\right)=11\)

Có \(4\) trường hợp : 

\(\Leftrightarrow\)\(\hept{\begin{cases}x+3=1\\x-7=11\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\x=18\end{cases}}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x+3=-1\\x-7=-11\end{cases}\Leftrightarrow\hept{\begin{cases}x=-4\\x=-4\end{cases}}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x+3=11\\x-7=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=8\\x=8\end{cases}}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x+3=-11\\x-7=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=-14\\x=6\end{cases}}}\)

Vậy \(x\in\left\{-4;8\right\}\)