Bài 2: 

a) Ta có: \(A=\left(7x+5\right)^2+\left(3x-5\right)^2-\left(10-6x\right)\left(5+7x\right)\)

\(=\left(7x+5\right)^2+2\cdot\left(7x+5\right)\cdot\left(3x-5\right)+\left(3x-5\right)^2\)

\(=\left(7x+5+3x-5\right)^2\)

\(=\left(10x\right)^2=100x^2\)

Thay x=-2 vào A, ta được:

\(A=100\cdot\left(-2\right)^2=100\cdot4=400\)

b) Ta có: \(B=\left(2x+y\right)\left(y^2-2xy+4x^2\right)-8x\left(x-1\right)\left(x+1\right)\)

\(=8x^3+y^3-8x\left(x^2-1\right)\)

\(=8x^3+y^3-8x^3+8x\)

\(=8x+y^3\)

Thay x=-2 và y=3 vào B, ta được:

\(B=-2\cdot8+3^3=-16+27=11\)

Ai help mk vs

Bài 1: 

Ta có: \(A=x\left(4-x\right)\)

\(=4x-x^2\)

\(=-\left(x^2-4x\right)\)

\(=-\left(x^2-4x+4\right)+4\)

\(=-\left(x-2\right)^2+4\le4\forall x\)

Dấu '=' xảy ra khi x=2

Vậy: \(A_{max}=4\) khi x=2

b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)

=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)

=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)

=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)

=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)

=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)

=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)

d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)

=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)

=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)

=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)

Bài 12:

a) \(\left(\dfrac{1}{2}x+4\right)^2\)

\(=\left(\dfrac{1}{2}x\right)^2+2\cdot\dfrac{1}{2}x\cdot4+4^2\)

\(=\dfrac{1}{4}x^2+4x+16\)

b) \(\left(7x-5y\right)^2\)

\(=\left(7x\right)^2-2\cdot7x\cdot5y+\left(5y\right)^2\)

\(=49x^2-70xy+25y^2\)

c) \(\left(6x^2+y^2\right)\left(y^2-6x^2\right)\)

\(=\left(y^2+6x^2\right)\left(y^2-6x^2\right)\)

\(=y^4-36x^4\)

d) \(\left(x+2y\right)^2\)

\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)

\(=x^2+4xy+4y^2\)

e) \(\left(x-3y\right)\left(x+3y\right)\)

\(=x^2-\left(3y\right)^2\)

\(=x^2-9y^2\)

f) \(\left(5-x\right)^2\)

\(=5^2-2\cdot5\cdot x+x^2\)

\(=25-10x+x^2\)

\(11,\)

\(a,\left(7x+4\right)^2-\left(7x+4\right)\left(7x-4\right)\)

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=\left(7x+4\right).8=56x+32\)

\(b,\left(x+2y\right)^2-6xy\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x+2y-6xy\right)\)

Bài `12`

`(1/2x+4)^2`

`=(1/2x)^2 + 2 . 1/2x.4 + 4^2`

`= 1/4 x^2 +4x + 16`

__

`(7x-5y)^2`

`=(7x)^2-2.7x.5y+(5y)^2`

`= 49x^2 -  70xy   + 25y^2`

__

`(6x^2+y^2)(y^2-6x^2)`

`=(y^2+6x^2)(y^2-6x^2)`

`=(y^2)^2 - (6x^2)^2`

`=y^4-36x^4`

__

`(x+2y)^2`

`=x^2+ 2.x.2y+(2y)^2`

`= x^2 + 4xy +4y^2`

__

`(x-3y)(x+3y)`

`=x^2 - (3y)^2`

`=x^2 - 9y^2`

__

`(5-x)^2`

`=5^2 -2.5.x+x^2`

`=25 - 10x+x^2`

Bài `11`

`(7x+4)^2 -(7x+4)(7x-4)`

`= (7x+4)(7x+4) -(7x+4)(7x-4)`

`=(7x+4)(7x+4-7x+4)`

`=8(7x+4)`

`= 56x+32`

__

`(x+2y)^2-6xy (x+2y)`

`= (x+2y) (x+2y-6xy)`

a) \(f\left(x\right)=5x^3-7x^2+2x+5\)

\(\Rightarrow f\left(1\right)=5.1^3-7.1^2+2.1+5\)

\(\Rightarrow f\left(1\right)=5.1-7.1+2+5\)

\(\Rightarrow f\left(1\right)=5-7+7\)

\(\Rightarrow f\left(1\right)=5\)

Vậy f(1) = 5.

\(g\left(x\right)=7x^3-7x^2+2x+5\)

\(\Rightarrow g\left(\frac{1}{2}\right)=7.\left(\frac{1}{2}\right)^3-7.\left(\frac{1}{2}\right)^2+2.\frac{1}{2}+5\)

\(\Leftrightarrow g\left(\frac{1}{2}\right)=7.\frac{1}{8}-7.\frac{1}{4}+1+5\)

\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{7}{8}-\frac{14}{8}+6\)

\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{-7}{8}+\frac{48}{8}\)

\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{41}{8}\)

Vậy \(g\left(\frac{1}{2}\right)=\frac{41}{8}\)

\(h\left(x\right)=2x^3+4x+1\)

\(\Rightarrow h\left(0\right)=2.0^3+4.0+1\)

\(\Rightarrow h\left(0\right)=0+0+1\)

\(\Rightarrow h\left(0\right)=1\)

Vậy \(h\left(0\right)=1\)

b)\(f\left(x\right)-g\left(x\right)+h\left(x\right)\)

\(=5x^3-7x^2+2x+5-2x^3-4x-1+7x^3-7x^2+2x+5\)

Rút gọn rồi tìm k(x)

Tìm M(x) tương tự

c) Bậc của k(x) là đơn thức có bậc cao nhất là 3

Nghiệm của k(x) là khi k(x) = 0 . Như câu a)

a) 4x²  - 2x + 3 - 4x.(x - 5) = 7x - 3

--> 4x² - 2x + 3 - 4x² + 20x = 7x - 3

--> 4x² - 2x - 4x² + 20x - 7x = -3 - 3

--> 11x = -6

--> x = \(\frac{-6}{11}\)

b) -3x.(x - 5) + 5.(x - 1) + 3x² = 4x

--> -3x² + 15x + 5x - 5 + 3x² = 4x

--> -3x²  + 15x + 5x + 3x² - 4x = 5 

--> 16x = 5

--> x = \(\frac{5}{16}\)

c) 7x.(x - 2) - 5.(x - 1) = 21x² - 14x² + 3

--> 7x² - 14x - 5x + 5 = 7x² + 3 

--> 7x²  - 14x - 5x - 7x²  = -5 + 3 

--> -19x = -2 

--> x = \(\frac{2}{19}\)

d) 3.(5x - 1) - x.(x - 2) + x² - 13x = 7

--> 15x - 3 - x² + 2x + x² - 13x = 7

--> 15x - x² + 2x + x² - 13x = 3 + 7

--> 4x = 10

--> x = \(\frac{5}{2}\)

e) \(\frac{1}{5}\)x.(10x - 15) - 2x.(x - 5) = 12

--> 2x² - 3x - 2x² + 10x = 12

--> 7x = 12

--> x = \(\frac{12}{7}\)

~ Học tốt ~

a) 4x² - 2x + 3 - 4x(x - 5) = 7x - 3

=> 4x² - 2x + 3 - 4x² + 20x = 7x - 3

=> 18x + 3 = 7x - 3

=> 18x - 7x = -3 - 3

=> 11x = -6

=>  x = -6/11

b) -3x(x - 5) + 5(x - 1) + 3x² = 4x

=> -3x² + 15x + 5x - 5 + 3x² = 4x

=> 20x - 5 = 4x

=> 20x - 4x = 5

=> 16x = 5

=> x = 5/16

\(c,7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)

\(\Leftrightarrow7x^2-14x-5x+5=7x^2+3\)

\(\Leftrightarrow7x^2-7x^2-19x=3-5\)

\(\Leftrightarrow-19x=-2\)

\(\Leftrightarrow x=\frac{2}{19}\)

a) 4x² - 2x + 3 - 4x.(x - 5) = 7x - 3

<=> 18x + 3 = 7x - 3

<=> 18x = 7x - 3 - 3

<=> 18x = 7x - 6

<=> 18x - 7x = -6

<=> 11x = -6

<=> x = -6/11

=> x = -6/11

b) -3x.(x - 5) + 5.(x - 1) + 3x² = 4x

<=> 20x - 5 = 4x

<=> 20x = 4x + 5

<=> 20x - 4x = 5

<=> 16x = 5

<=> x = 5/16

=> x = 5/16

c) 7x.(x - 2) - 5.(x - 1) = 21x² - 14x² + 3

<=> 7x.(x - 2) - 5.(x - 1) = 7x² + 3

<=> 7x² - 19x + 5 = 7x² + 3

<=> 7x² - 19x = 7x² + 3 - 5

<=> 7x² - 19x = 7x² - 2 

<=> 7x² - 19x - 7x² = -2

<=> -19x = -2

<=> x = 2/19

=> x = 2/19

d) 3.(5x - 1) - x.(x - 2) + x² - 13x = 7

<=> 4x - 3 = 7

<=> 4x = 7 + 3

<=> 4x = 10

<=> x = 10/4

=> x = 5/2

e) 1/5x.(10x - 15) - 2x.(x - 5) = 12

<=> x(2x - 3) - 2x(x - 5) = 12

<=> 7x = 12

<=> x = 12/7

=> x = 12/7

a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)

Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)

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

Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)

\(\Leftrightarrow2x^2+2-2x^2-2x=0\)

\(\Leftrightarrow-2x+2=0\)

\(\Leftrightarrow-2x=-2\)

hay x=1(nhận)

Vậy: S={1}

b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)

Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)

\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)

\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)

\(\Leftrightarrow-56x-1=0\)

\(\Leftrightarrow-56x=1\)

hay \(x=-\dfrac{1}{56}\)(nhận)

Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)

Ta có: \(\dfrac{5}{3x+2}=2x-1\)

\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)

\(\Leftrightarrow6x^2-3x+4x-2-5=0\)

\(\Leftrightarrow6x^2+x-7=0\)

\(\Leftrightarrow6x^2-6x+7x-7=0\)

\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)

d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)

Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)

\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)

\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)

f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)