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

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

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

\(=\dfrac{6x^2+2}{\left(x-1\right)\left(x+1\right)}\)

giúp e phần C vs ạ

Bài 8 

a, \(A=a^2+b^2=\left(a+b\right)^2-2ab\Rightarrow S^2-2P\)

b, \(B=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]\)

\(\Rightarrow S\left(S^2-3P\right)=S^3-3PS\)

c, \(C=a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=\left[\left(a+b\right)^2-2ab\right]^2-2\left(ab\right)^2\)

\(\Rightarrow\left(S^2-2P\right)^2-2P^2\)

có thể giúp mik các bài còn lại đc ko

Câu 1:

a: 5x-2=3x+6

=>5x-3x=2+6

=>2x=8

=>\(x=\dfrac{8}{2}=4\)

b: a<=b

=>-2022a>=-2022b

=>-2022a+2021>=-2022b+2021

Câu 2:

1:

a: ĐKXĐ: x<>1

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

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

=>\(2x+5=x+2\)

=>x=-3(nhận)

b: |x-9|=2x-3

=>\(\left\{{}\begin{matrix}2x-3>=0\\\left(2x-3\right)^2=\left(x-9\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\\left(2x-3-x+9\right)\left(2x+3+x-9\right)=0\end{matrix}\right.\)

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

=>x=2

2:

\(\dfrac{x-3}{2}-\dfrac{3x+2}{4}< \dfrac{1}{3}\)

=>\(\dfrac{6\left(x-3\right)-3\left(3x+2\right)}{12}< \dfrac{4}{12}\)

=>6x-18-9x-6<4

=>-3x-24<4

=>-3x<28

=>\(x>-\dfrac{28}{3}\)

Câu 3:

Gọi độ dài quãng đường AB là x(km)

(Điều kiện: x>0)

Thời gian đi từ A đến B là \(\dfrac{x}{40}\left(giờ\right)\)

Thời gian ô tô đi từ B về A là \(\dfrac{x}{30}\left(giờ\right)\)

Theo đề, ta có phương trình:

\(\dfrac{x}{40}+\dfrac{x}{30}+\dfrac{1}{2}=9+\dfrac{1}{4}\)

=>\(\dfrac{7x}{120}=8,75\)

=>\(x=8,75:\dfrac{7}{120}=120\cdot1,25=150\left(nhận\right)\)

vậy: Độ dài quãng đường AB là 150km

1/ Xét \(\Delta ABC\) và \(\Delta HAC\) có:

∠A = ∠AHC = 90 độ

∠C là góc chung

Do đó: △ABC ∼ △HAC (g . g)

2/ Ta có: \(\Delta HAC\sim\Delta ABC\)

\(\Rightarrow\dfrac{AC}{HC}=\dfrac{BC}{AC}\)

\(\Rightarrow AC.AC=HC.BC\)

\(\Rightarrow AC^2=HC.BC\) (đpcm)

3/ Đặt BD là x, theo tính chất đường phân giác, ta có:

\(\dfrac{AB}{AC}=\dfrac{BD}{DC}\Rightarrow\dfrac{9}{12}=\dfrac{x}{15-x}\)

\(\Rightarrow9\left(15-x\right)=12x\)

\(\Rightarrow135-9x=12x\)

\(\Rightarrow135=12x+9x\)

\(\Rightarrow135=21x\)

\(\Rightarrow x\approx6,4\)

Độ dài của DC là: \(15-x\Rightarrow15-6,4=8,6\)

Vậy BD = 6,4 cm và DC = 8,6 cm

A)x=10cm

B)x=11cm

Câu 3: 

a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)

\(=6x^2-2x-6x^2-2x+18x+6\)

=14x+6

b: Ta có: \(2x\left(x+7\right)-3x\left(x+1\right)\)

\(=2x^2+14x-3x^2-3x\)

\(=-x^2+11x\)

Câu 2: 

a: Ta có: \(\left(-8x^5+12x^3-16x^2\right):4x^2\)

\(=-8x^5:4x^2+12x^3:4x^2-16x^2:4x^2\)

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

b: Ta có: \(\left(12x^3y^3-18x^2y+9xy^2\right):6xy\)

\(=12x^3y^3:6xy-18x^2y:6xy+9xy^2:6xy\)

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

c: Ta có: \(\dfrac{x^3-11x^2+27x-9}{x-3}\)

\(=\dfrac{x^3-3x^2-8x^2+24x+3x-9}{x-3}\)

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

d: Ta có: \(\dfrac{6x^4-13x^3+7x^2-x-5}{3x+1}\)

\(=\dfrac{6x^4+2x^3-15x^3-5x^2+12x^2+4x-5x-\dfrac{5}{3}-\dfrac{10}{3}}{3x+1}\)

\(=2x^3-5x^2+4x-\dfrac{5}{3}-\dfrac{\dfrac{10}{3}}{3x+1}\)

Câu 5:

a. $|x+\frac{4}{5}|-\frac{1}{7}=0$

$|x+\frac{4}{5}|=\frac{1}{7}$

$\Rightarrow x+\frac{4}{5}=\pm \frac{1}{7}$

$\Rightarrow x=\frac{-23}{35}$ hoặc $x=\frac{-33}{35}$

v.

$2x+5-(x-7)=18$

$2x+5-x+7=18$

$x+12=18$

$x=6$

c.

$2(x+1)+4^2=2^4$
$2(x+1)+16=16$

$2(x+1)=0$

$x+1=0$

$x=-1$

d.

$\frac{x-3}{x+5}=\frac{5}{7}$

$\Rightarrow 7(x-3)=5(x+5)$

$\Rightarrow 7x-21=5x+25$

$\Rightarrow 2x=46$

$\Rightarrow x=23$

Câu 5:

\(a,\left|x+\dfrac{4}{5}\right|-\dfrac{1}{7}=0\\ \Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{7}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{7}\\x+\dfrac{4}{5}=-\dfrac{1}{7}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}-\dfrac{4}{5}\\x=-\dfrac{1}{7}-\dfrac{4}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{35}\\x=-\dfrac{33}{35}\end{matrix}\right.\\ b,2x+5-\left(x-7\right)=18\\ \Leftrightarrow2x-x=18-5-7\\ \Leftrightarrow x=6\\ c,2\left(x+1\right)+4^2=2^4\\ \Leftrightarrow2\left(x+1\right)=2^4-4^2=16-16\\ \Leftrightarrow2\left(x+1\right)=0\\ \Rightarrow x+1=0\\ \Leftrightarrow x=0-1=-1\\ d,\dfrac{x-3}{x+5}=\dfrac{5}{7}\left(x\ne-5\right)\\ \Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\\ \Leftrightarrow7x-21=5x+25\\ \Leftrightarrow7x-5x=25+21\\ \Leftrightarrow2x=46\\ \Leftrightarrow x=23\)