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AH
Akai Haruma
Giáo viên
10 tháng 8 2023

Lời giải:

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

$\Leftrightarrow x^2-25-(x^2+6x+9)+3(x^2-4x+4)=(x^2+2x+1)-(x^2-16)+3x^2$
$\Leftrightarrow 3x^2-18x-22=3x^2+2x+17$

$\Leftrightarrow -18x-22=2x+17$

$\Leftrightarrow 20x=-39$

$\Leftrightarrow x=\frac{-39}{20}$

28 tháng 11 2016

làm nốt

d) (2x-1)(3x+2)(3-x)

=(6x2+x-2)(3-x)

=-6x3+17x2+5x-6

e) (x+3)(x2+3x-5)

=x3+6x2+4x-15

f) (xy-2)(x3-2x-6)

=x4y-2x3-2x2y-6xy+4x+12

g) (5x3-x2+2x-3)(4x2-x+2)

=20x5-9x4+19x3-16x2+7x-6

 

28 tháng 11 2016

Bài 1:

a) (x-2)(x2+3x+4)

=x(5x+4)-2(5x+4)

= 5x2+4x-10x-8

=5x2-6x-8

16 tháng 9 2018

e, (x-1)(x2 + x + 1)-x(x+2)(x-2) = 5

x(x2 +x + 1 ) - (x2 + x +1 )- [ x (x2 - 4)] = 5

x3 +x2 +x - x2 - x - 1 - x3 +4x = 5

4x - 1 = 5

4x = 6

x =\(\dfrac{3}{2}\)

f, (x-1)3 - (x+3)(x2 - 3x +9 ) +3(x2 - 4) = 2

x - 3x2 +3x - 1 - [( x3 - 3x2 + 9x) + (3x2 - 9x +27)] = 2

x3 - 3x2 + 3x - 1 -x3 +3x2 -9x - 3x2 +9x - 27 +3x2 - 12 = 2

3x - 1 - 27 - 12 = 2

3x = 42

x = 14

16 tháng 9 2018

muốn tao trả lờ cho ko , mai đến lớp nhá

11 tháng 8 2016

Bài 1:

a. A = x^2 - 5x - 1

\(=x^2-5x+\frac{25}{4}-\frac{29}{4}\)

\(=x^2-5x+\left(\frac{5}{2}\right)^2-\frac{29}{4}\)

\(=\left(x-\frac{5}{2}\right)^2-\frac{29}{4}\ge0-\frac{29}{4}=-\frac{29}{4}\)

Dấu = khi x=5/2

Vậy MinC=-29/4 khi x=5/2

 

 

11 tháng 8 2016

2. Tìm x:
a. ( 2x - 3 )^2 - ( 4x + 1 )( 4x - 1 ) = ( 2x - 1 ).( 3 - 7x )

=>4x2-12x+9+1-16x2=-14x2+13x-3

=>-12x2-12x+10=-14x2+13x-3

=>2x2-25x+13=0

\(\Rightarrow2\left(x-\frac{25}{4}\right)^2-\frac{521}{8}=0\)

\(\Rightarrow\left(x-\frac{25}{4}\right)^2=\frac{521}{16}\)

\(\Rightarrow x-\frac{25}{4}=\pm\sqrt{\frac{521}{16}}\)

\(\Rightarrow x=\frac{25}{4}\pm\frac{\sqrt{521}}{4}\)

c. 4.( x - 3 ) - ( x + 2 ) = 0

=>4x-12-x-2=0

=>3x-14=0

=>3x=14

=>x=14/3

 

 

11 tháng 11 2018

Hỏi đáp Toán

11 tháng 11 2018

Hỏi đáp Toán

4 tháng 12 2018

a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)

\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)

\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)

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

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

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

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

\(=\left(x^2+2\right)^2-\left(x^4-16\right)\)

\(=\left(x^4+4x^2+4\right)-\left(x^4-16\right)\)

\(=x^4+4x^2+4-x^4+16\)

\(=4x^2+20\)

\(=4\left(x^2+5\right)\)

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

\(=\left(4x^2-4xy+y^2\right)-2\left(x^2+6xy+9y^2\right)-\left(9x^2-1\right)\)

\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)

\(=-7x^2-20xy-17y^2+1\)

d) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)

\(=\left(x^6-3x^4+3x^2-1\right)-\left(x^6-1\right)\)

\(=x^6-3x^4+3x^2-1-x^6+1\)

\(=-3x^4+3x^2\)

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

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

e) \(\left(2x-1\right)^2-2\left(4x^2-1\right)+\left(2x+1\right)^2\)

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

\(=\left[\left(2x-1\right)-\left(2x+1\right)\right]^2\)

\(=\left(2x-1-2x-1\right)^2\)

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

g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)

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

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

\(=\left(x+2z\right)^2\)

h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)

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

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

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

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

i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)

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

\(=5x^2-\left(-2x^2-3x+1\right)-3\left(x+1\right)\)

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

\(=7x^2-4\)

30 tháng 3 2018

Hỏi đáp Toán

30 tháng 3 2018

Dài quá c ơi :<

8 tháng 1 2018

Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)

14 tháng 2 2020

a.=\(\frac{7x+2}{3xy^2}.\frac{x^2y}{14x+4}\)

=\(\frac{7x+2}{3y}.\frac{x^2y}{2\left(7x+2\right)}\)

=\(\frac{1}{3y}.\frac{x}{2}\)

=\(\frac{x}{6y}\)

b.=\(\frac{8xy}{3x-1}.\frac{5-15x}{12xy^3}\)

=\(\frac{2}{3x-1}.\frac{-15x+5}{3y^2}\)

=\(\frac{2}{3x-1}.\frac{-5\left(3x-1\right)}{3y^2}\)

=\(\frac{-10}{3y^2}\)

c.=\(\frac{3\left(x^3+1\right)}{x-1}.\frac{1}{x^2-x+1}\)

=\(\frac{3\left(x+1\right).\left(x^2-x+1\right)}{x-1}.\frac{1}{x^2-x+1}\)

=\(\frac{3x+3}{x-1}\)

d.=\(\frac{4\left(x+3\right)}{.\left(3x-1\right)}.\frac{1-3x}{x^2+3x}\)

=\(\frac{4\left(x+3\right)}{x.\left(3x-1\right)}.\frac{-\left(3x-1\right)}{x\left(x+3\right)}\)

=\(\frac{-4}{x^2}\)

e.=\(\frac{2\left(2x+3y\right)}{x-1}.\frac{1-x^3}{4x^2+12xy+9y^2}\)

=\(2.\frac{-\left(1+x+x^2\right)}{2x+3y}\)

=\(-\frac{2x^2+2x+2}{2x+3y}\)

14 tháng 2 2020

Phần C thiếu x3 , chỗ (x-1)