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22 tháng 7 2023

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

\(=x^3+1\)

b) \(\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

\(=8x^3-\dfrac{1}{8}\)

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

`b, (2x-1/2)(4x^2 + x + 1/4) = 8x^3 - 1/8`

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

b: \(=\dfrac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)}{x\left(x-1\right)}=\dfrac{-2\left(x+1\right)}{\left(x-1\right)\left(x+2\right)}\)

a: ta có: \(\dfrac{\left(x+2\right)^2}{2}+\dfrac{\left(2x+1\right)^2}{4}+\dfrac{\left(2x-1\right)^2}{8}-\left(x+1\right)^2=0\)

\(\Leftrightarrow4\left(x^2+4x+4\right)+2\left(4x^2+4x+1\right)+4x^2-4x+1-8\left(x+1\right)^2=0\)

\(\Leftrightarrow4x^2+16x+16+8x^2+8x+2+4x^2-4x+1-8\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow16x^2+20x+19-8x^2-16x-8=0\)

\(\Leftrightarrow8x^2+4x+11=0\)

\(\text{Δ}=4^2-4\cdot8\cdot11=-336< 0\)

Vì Δ<0 nên phương trình vô nghiệm

AH
Akai Haruma
Giáo viên
13 tháng 9 2021

b.

PT \(\Leftrightarrow \frac{x^2+2x+1}{2}-\frac{4x^2-4x+1}{3}+\frac{4x^2+4x+1}{4}-\frac{x^2-10x+25}{6}=0\)

\(\Leftrightarrow \left(\frac{x^2+2x+1}{2}+\frac{4x^2+4x+1}{4}\right)-\left(\frac{4x^2-4x+1}{3}+\frac{x^2-10x+25}{6}\right)=0\)

\(\Leftrightarrow \frac{6x^2+8x+3}{4}-\frac{9x^2-18x+27}{6}=0\)

\(\Leftrightarrow \frac{3(6x^2+8x+3)-2(9x^2-18x+27)}{12}=0\)

$\Leftrightarrow 5x-\frac{15}{4}=0$

$\Leftrightarrow x=\frac{3}{4}$

 

b)

ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)

Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)

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

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

Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Leftrightarrow2x^2-14=2x^2+x-10\)

\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)

\(\Leftrightarrow-x-4=0\)

\(\Leftrightarrow-x=4\)

hay x=-4(nhận)

Vậy: S={-4}

22 tháng 12 2020

a) Ta có: \(\dfrac{9-3x}{x^2+3x+4}-\dfrac{3x-23}{\left(1-x\right)\left(x+4\right)}\)

\(=\dfrac{9-3x}{x^2+3x+4}+\dfrac{3x-23}{x^2+3x-4}\)

\(=\dfrac{\left(9-3x\right)\left(x^2+3x-4\right)}{\left(x^2+3x+4\right)\left(x^2+3x-4\right)}+\dfrac{\left(3x-23\right)\left(x^2+3x+4\right)}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)

\(=\dfrac{9x^2+27x-36-3x^3-9x^2+12x+3x^3+9x^2+12x-23x^2-69x-92}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)

\(=\dfrac{-14x^2-18x-128}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)

b) Ta có: \(\dfrac{4-x}{x^3+2x}-\dfrac{x+5}{x^3-x^2+2x-2}\)

\(=\dfrac{4-x}{x\left(x^2+2\right)}-\dfrac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)

\(=\dfrac{4-x}{x\left(x^2+2\right)}-\dfrac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)

\(=\dfrac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\dfrac{x\left(x+5\right)}{x\left(x-1\right)\left(x^2+2\right)}\)

\(=\dfrac{4x-4-x^2+x-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)

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

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

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

 

AH
Akai Haruma
Giáo viên
18 tháng 12 2023

Bài 1:

a. $2x^3+3x^2-2x=2x(x^2+3x-2)=2x[(x^2-2x)+(x-2)]$

$=2x[x(x-2)+(x-2)]=2x(x-2)(x+1)$

b.

$(x+1)(x+2)(x+3)(x+4)-24$

$=[(x+1)(x+4)][(x+2)(x+3)]-24$

$=(x^2+5x+4)(x^2+5x+6)-24$

$=a(a+2)-24$ (đặt $x^2+5x+4=a$)

$=a^2+2a-24=(a^2-4a)+(6a-24)$

$=a(a-4)+6(a-4)=(a-4)(a+6)=(x^2+5x)(x^2+5x+10)$

$=x(x+5)(x^2+5x+10)$

AH
Akai Haruma
Giáo viên
18 tháng 12 2023

Bài 2:

a. ĐKXĐ: $x\neq 3; 4$

\(A=\frac{2x+1-(x+3)(x-3)+(2x-1)(x-4)}{(x-3)(x-4)}\\ =\frac{2x+1-(x^2-9)+(2x^2-9x+4)}{(x-3)(x-4)}\\ =\frac{x^2-7x+14}{(x-3)(x-4)}\)

b. $x^2+20=9x$

$\Leftrightarrow x^2-9x+20=0$

$\Leftrightarrow (x-4)(x-5)=0$

$\Rightarrow x=5$ (do $x\neq 4$)

Khi đó: $A=\frac{5^2-7.5+14}{(5-4)(5-3)}=2$

8 tháng 12 2021

\(a,=\dfrac{x^2-20+x^2-7x+10+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\\ b,=\dfrac{10x+15-4x+6+2x-9}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{4}{2x-3}\\ c,=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\\ =\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{x+4-x}{x\left(x+4\right)}=\dfrac{4}{x\left(x+4\right)}\)

26 tháng 12 2021

a: \(=\dfrac{x-z}{2}\)

b: \(=\dfrac{3x}{4y^3}\)