e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

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

f: Ta có: \(2x^2+7x+3=0\)

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

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

5x-16=40+x

=> 5x-16-x = 40

=> 5x-x -16=40

4x-16=40

4x= 40+16

4x=56

x= 56:4

x=14

Vậy...

4x-10=15-x

=> 4x-10+x= 15

4x+x -10=15

5x= 15+10

5x= 25

x= 25:5

x=5

Vậy....

5x -16=40+x

=> 5x-x=40+16

=>4x=56

=>x=56:4

x=14

a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)

b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)

c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Xem lại đề câu d 

Các bn giúp mình nhanh đk ko. Cảm ơn các bạn nhiều nha.

bạn cứ chia ra trường hợp là đc

a: =>-140x+1050-132+84x=100

=>-56x=100+132-1050=-818

=>x=409/28

b: \(\Leftrightarrow82-287x-15x+60=20\)

=>-302x+142=20

=>-302x=-122

=>x=61/151

c: =>165x-66+12x-32=15

=>177x=113

=>x=113/177

d: =>140-16x-77+22x=19

=>6x+63=19

=>6x=-44

=>x=-22/3

a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 40x = 15 <=> x = 3/8

a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 5(8x-5) = 15

<=> 40x = 40 <=> x = 1

Cái này mới chuẩn

b) (2x+1)(1-2x)+(1-2x)²=18 <=> 1 - 4x² + 4x² - 4x + 1 = 18

<=> -4x = 16 <=> x = -4

24: 

\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)

\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)

=>(x+10)(x-2)=0

=>x=-10 hoặc x=2

25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)

\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)

\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)

\(\Leftrightarrow x+5=0\)

hay x=-5

a/\(7x^2+8x-15=0\Leftrightarrow\left(x-1\right)\left(7x+15\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-15}{7}\end{cases}}\)

b/\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)\(\Leftrightarrow4x^2-25-4x^2-4x+35=0\)

\(\Leftrightarrow-4x+10=0\Leftrightarrow x=\frac{5}{2}\)

a) 7x² + 8x - 15 = 0

giải phương trình trên máy tính ta có :

x = 1  hoặc x = \(-\frac{15}{7}\)

b) Viết thiếu đề nhưng tôi cứ cho nó = 0 nếu là kết quả khác thì tham khảo 

4x² - 25 - ( 2x - 5 ) (2x + 7 ) = 0

=> 4x² - 25 - 4x² - 4x + 35 = 0

=> -4x + 10 = 0

=>  - 4x = - 10 

=> x = \(\frac{5}{2}\)

Study well 

a) \(7x^2+8x-15=0\)

\(\cdot\Delta=8^2-4.7.\left(-15\right)=484,\sqrt{\Delta}=22\)

\(\cdot\)\(\Delta>0\)nên pt có 2 nghiệm phân biệt

\(x_1=\frac{-8+22}{14}=1\); \(x_1=\frac{-8-22}{14}=\frac{-15}{7}\)

\(x=-\dfrac{1}{2}=-0.5,y=\dfrac{5}{4}=1.25\\x=2,y=\dfrac{7}{2}=3.5\)