Đặt 2x²+x-2015=a; x²-5x-2016=b

phương trình tương đương a²+4b²=4ab

=> a²-4ab+4b²=0

=> (a-2b)²=0

=> a=2b

vậy 2x²+x-2015=2*(x²-5x-2016)

=> x=\(\frac{-2017}{11}\)

1.x²-9

= (x-3)(x+3)

2. -2x²+2x+12

= -2x²+6x-4x+12

= -2x(x+2)+6(x+2)

= (x+2)(-2x+6)

4. -2x²+2x+24

= -2x²+8x-6x+24

= -2x(x+3)+8(x+3)

= (x+3)(-2x+8)

6. x²-5x+4

= x²-4x-x+4

= x(x-1) -4(x-1)

= (x-1)(x-4)

8. x²-7x+6

= x²-6x-x+6

= x(x-1)-6(x-1)

= (x-1)(x-6)

9. x²+5x+4

= x²+4x+x+4

= x(x+1)+4(x+1)

=(x+1)(x+4)

10. x²+7x+6

= x² +x+6x+6

= x(x+1)+6(x+1)

= (x+6)(x+1)

K nhé

Cảm ơn nhìu

7. x³+x(3xy+3xz+2yz)+y³+y(3yx+3yz+2xz)+z³+z(3zx+3zy+2xy)

= x³+3x²y+3x²z+2xyz+y³+3y²x+3y²z+2xyz+z³+3z²x+3z²y+2xyz

= (x³+3x²y+3xy²+y³)+z³+(3x²z+6xyz+3y²z)+3xz²+3yz²

= (x+y)³+z³+3(x+y)²z+3(x+y)z²

= (x+y+z)³

Ta có : \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\)

\(\Rightarrow\frac{x+2015}{5}+\frac{x+2016}{6}-\frac{x+2017}{7}-\frac{x+2018}{8}=0\)

\(\Rightarrow\left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)-\left(\frac{x+2017}{7}-1\right)-\left(\frac{x+2018}{8}-1\right)=0\)

\(\Leftrightarrow\frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\)

Mà \(\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)\ne0\)

\(\Rightarrow x+2010=0\)

=> x = -2010

\(\left(-x^3.z.y\right).\left(\dfrac{2}{3}.y.x^2\right)^2\)

\(=-x^3.z.y.\dfrac{4}{9}.y^2.x^4\)

\(=-\dfrac{4}{9}x^7.y^3.z\)

`(-x^3zy)(2/3yx^2)^2`

`=-4/9x^3zy.y^2x^4`

`=-4/9x^{3+4}.y^{1+2}z`

`=-4/9x^7y^3z`

\(\left(-x^3zy\right)\left(\dfrac{2}{3}yx^2\right)^2\)

\(\Leftrightarrow\left(-x^3yz\right)\left(\dfrac{2}{3}x^2y\right)\left(\dfrac{2}{3}x^2y\right)\)

\(\Leftrightarrow-\left[x^7y^3z\left(\dfrac{2}{3}\right)^2\right]\)

\(\Leftrightarrow-\dfrac{4}{9}x^7y^3z\)

a) I2x-5I=13

<=> 2x-5=13 hoặc 2x-5=-13

<=> 2x=18 hoặc 2x=-8

<=> x=9 hoặc x=-4

b) I7x+3I=66

<=> 7x+3=66 hoặc 7x+3=-66

<=> 7x=63 hoặc 7x=-69

<=> x=9 hoặc x=\(\frac{-69}{7}\)

a) \(\left|2x-5\right|=13\)

\(\Leftrightarrow2x-5=\pm13\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=13\\2x-5=-13\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=18\\2x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\2x=-4\end{cases}}}\)

Vậy \(x\in\left\{9;-4\right\}\)

b) \(\left|7x+3\right|=66\)

\(\Leftrightarrow7x+3=\pm66\)

\(\Leftrightarrow\orbr{\begin{cases}7x+3=66\\7x+3=-66\end{cases}\Leftrightarrow\orbr{\begin{cases}7x=63\\7x=-69\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\x=\frac{-69}{7}\end{cases}}}\)

Vậy \(x\in\left\{9;\frac{-69}{7}\right\}\)

a . | 2x - 5 | = 13

\(\Rightarrow\orbr{\begin{cases}2x-5=13\\2x-5=-13\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=13+5\Rightarrow x=18:2=9\\2x=-13+5\Rightarrow x=-8:2=-4\end{cases}}\)

Vậy .....................................................................

b. | 7x + 3 | = 66

 \(\Rightarrow\orbr{\begin{cases}7x+3=66\\7x+3=-66\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}7x=63\Rightarrow x=63:7=9\\7x=-69\Rightarrow x=-69:7=-\frac{69}{7}\end{cases}}\)

vậy....................................................................................

Sửa đề:

\((2x^2+x-2015)^2+4(x^2-5x-2016)^2=4(2x^2+x-2015)(x^2-5x-2016)\)

\(\Rightarrow\left(2x^2+x-2015\right)^2-2.\left(2x^2+x-2015\right).2.\left(x^2-5x-2016\right)+[2.\left(x^2-5x-2016\right)]^2=0\)

\(\Rightarrow[2x^2+x-2015-2.\left(x^2-5x-2016\right)]^2=0\)

\(\Rightarrow11x+2017=0\)

\(\Rightarrow x=\frac{-2017}{11}\)

a.\(=\dfrac{8}{15}\times\dfrac{1}{2}=\dfrac{4}{15}\)

b.\(=\dfrac{1}{3}\times\dfrac{1}{2}-\dfrac{1}{5}\times\dfrac{1}{2}=\dfrac{1}{6}-\dfrac{1}{10}=\dfrac{1}{15}\)

a.(1/3+1/5) x 1/2

=8/15 x 1/2=8/30=4/15

b)(1/3-1/5) x 1/2=2/15 x 1/2=1/15