`7,`

`a, B+A=4x-2x^2+3`

`-> B=(4x-2x^2+3)-A`

`-> B=(4x-2x^2+3)-(x^2-2x+1)`

`B=4x-2x^2+3-x^2+2x-1`

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

`B=-3x^2+6x+2`

`b, C-A=-x+7`

`-> C=(-x+7)+A`

`-> C=(-x+7)+(x^2-2x+1)`

`-> C=-x+7+x^2-2x+1`

`C=x^2+(-x-2x)+(7+1)`

`C=x^2-3x+8`

`c,`

`A-D=x^2-2`

`-> D= A- (x^2-2)`

`-> D=(x^2-2x+1)-(x^2-2)`

`D=x^2-2x+1-x^2+2`

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

`D=-2x+3`

`6,`

`a,`

`P+Q=4x-2x^2+3`

`-> Q=(4x-2x^2+3)-P`

`-> Q=(4x-2x^2+3)-(3x^2+x-1)`

`Q=4x-2x^2+3-3x^2-x+1`

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

`Q=x^2+3x+4`

`b,`

`x^2-5x+2-P=H`

`-> H= (x^2-5x+2)-(3x^2+x-1)`

`H=x^2-5x+2-3x^2-x+1`

`H=(x^2-3x^2)+(-5x-x)+(2+1)`

`H=-4x^2-6x+3`

`c,`

`P-R=5x^2-3x-4`

`-> R= P- (5x^2-3x-4)`

`-> R=(3x^2+x-1)-(5x^2-3x-4)`

`R=3x^2+x-1-5x^2+3x+4`

`R=(3x^2-5x^2)+(x+3x)+(-1+4)`

`R=-2x^2+4x+3`

hong bít

`a)2x^2+3(x-1)(x+1)=5x(x+1)`

`<=>2x^2+3x^2-3=5x^2+5x`

`<=>5x=-3`

`<=>x=-3/5`

__________________________________________

`b)(x-3)^3+3-x=0` nhỉ?

`<=>(x-3)^3-(x-3)=0`

`<=>(x-3)(x^2-1)=0`

`<=>[(x=3),(x^2=1<=>x=+-1):}`

__________________________________________

`c)5x(x-2000)-x+2000=0`

`<=>5x(x-2000)-(x-2000)=0`

`<=>(x-2000)(5x-1)=0`

`<=>[(x=2000),(x=1/5):}`

__________________________________________

`d)3(2x-3)+2(2-x)=-3`

`<=>6x-9+4-2x=-3`

`<=>4x=2`

`<=>x=1/2`

__________________________________________

`e)x+6x^2=0`

`<=>x(1+6x)=0`

`<=>[(x=0),(x=-1/6):}`

\(y'=\dfrac{\left(-2x+2\right)\left(x-3\right)-\left(-x^2+2x+c\right)}{\left(x-3\right)^2}=\dfrac{-x^2+6x-6-c}{\left(x-3\right)^2}\)

\(\Rightarrow\) Cực đại và cực tiểu của hàm là nghiệm của: \(-x^2+6x-6-c=0\) (1)

\(\Delta'=9-\left(6+c\right)>0\Rightarrow c< 3\)

Gọi \(x_1;x_2\) là 2 nghiệm của (1) \(\Rightarrow\left\{{}\begin{matrix}-x_1^2+6x_1-6=c\\-x_2^2+6x_2-6=c\end{matrix}\right.\)

\(\Rightarrow m-M=\dfrac{-x_1^2+2x_1+c}{x_1-3}-\dfrac{-x_2^2+2x_2+c}{x_2-3}=4\)

\(\Leftrightarrow\dfrac{-2x_1^2+8x_1-6}{x_1-3}-\dfrac{-2x_2^2+8x_2-6}{x_2-3}=4\)

\(\Leftrightarrow2\left(1-x_1\right)-2\left(1-x_2\right)=4\)

\(\Leftrightarrow x_2-x_1=2\)

Kết hợp với Viet: \(\left\{{}\begin{matrix}x_2-x_1=2\\x_1+x_2=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=2\\x_2=4\end{matrix}\right.\)

\(\Rightarrow c=2\)

Có 1 giá trị nguyên

`a)25/(x+1)-1 1/6=-1/3-0,5`

`=>25/(x+1)=-1/3-1/2+1+1/6`

`=>25/(x+1)=1/3`

`=>75=x+1`

`=>x=74`

Vậy `x=74`

`b)(2x+25 3/5)^2-9/25=0`

`=>(2x+128/5)=9/25`

`**2x+128/5=3/5`

`=>2x=-125/5=-25`

`=>x=-25/2`

`**2x+128/5=-3/5`

`=>2x=-131/5`

`=>x=-131/10`

Giải:

a) \(\dfrac{25}{x+1}-1\dfrac{1}{6}=\dfrac{-1}{3}-0,5\) 

              \(\dfrac{25}{x+1}=\dfrac{-5}{6}+\dfrac{7}{6}\) 

              \(\dfrac{25}{x+1}=\dfrac{1}{3}\) 

\(\Rightarrow1.\left(x+1\right)=25.3\)  

\(\Rightarrow x+1=75\) 

\(\Rightarrow x=75-1\) 

\(\Rightarrow x=74\) 

b) \(\left(2x+25\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\) 

              \(\left(2x+\dfrac{128}{5}\right)^2=0+\dfrac{9}{25}\) 

             \(\left(2x+\dfrac{128}{5}\right)^2=\dfrac{9}{25}\) 

\(\Rightarrow\left[{}\begin{matrix}\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{128}{5}=\dfrac{3}{5}\\2x+\dfrac{128}{5}=\dfrac{-3}{5}\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-25}{2}\\x=\dfrac{-131}{10}\end{matrix}\right.\) 

Chúc bạn học tốt!

CM đẳng thức hay tìm x,y vậy 

Mình sẽ làm theo đề bài của mình nếu đúng thì ... nha 

Biến đổi vế phải  ta có :

( x + y) [ ( x - y)^2 + xy ] = ( x + y)( x^2 - 2xy + y^2 + xy)

                                      = ( x+  y)( x^2 - xy+ y^2)

                                       = x^3 + y^3

VẬy VT  = VP đẳng thức được CM 

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