a) A = 3 ( x − y ) 2 − 2 ( x + y ) 2 − ( x − y ) ( x + y ) 2 A = [ ( x − y ) − ( x + y ) ] 2 + 5 ( x − y ) 2 − 5 ( x + y ) 2 2 A = 4 y 2 + 5 [ ( x − y ) − ( x + y ) ] [ ( x − y ) + ( x + y ) ] 2 A = 4 y 2 + 5 [ − 2 y ] [ 2 x ] = 4 y 2 − 20 x y = 4 y ( y − 5 x ) A = 2 y ( y − 5 x )

bạn ko hiểu thì đây nhé

Giúp mk vs Akai HarumaLovers

a) \(x-\dfrac{3}{4}=6\times\dfrac{3}{8}\)

\(x-\dfrac{3}{4}=\dfrac{9}{4}\)

=> \(x=\dfrac{9}{4}+\dfrac{3}{4}=3\)

b) \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)

\(\dfrac{7}{8}:x=\dfrac{5}{2}\)

=> \(x=\dfrac{7}{8}:\dfrac{5}{2}=\dfrac{7}{20}\)

c) \(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)

\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

=> \(x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)

d) \(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)

\(\dfrac{6}{5}-x=\dfrac{2}{3}\)

=> \(x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{8}{15}\)

e) \(x\times3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\)(?)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

=> \(x=\dfrac{40}{51}:\dfrac{10}{3}=\dfrac{4}{17}\)

f) \(5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\)

\(\dfrac{17}{3}:x=\dfrac{5}{3}\)

=> \(x=\dfrac{17}{3}:\dfrac{5}{3}=\dfrac{17}{5}\)

a: =>x-3/4=18/8=9/4

=>x=9/4+3/4=12/4=3

b: =>7/8:x=5/2

=>x=7/8:5/2=7/8*2/5=14/40=7/20

c: x+1/2*1/3=3/4

=>x+1/6=3/4

=>x=3/4-1/6=9/12-2/12=7/12

d: =>12/10-x=2/3

=>6/5-x=2/3

=>x=6/5-2/3=18/15-10/15=8/15

e: =>x*10/3=10/3:17/4=10/3*4/17

=>x=4/17

f: =>17/3:x=13/3-5/2=26/6-15/6=11/6

=>x=17/3:11/6=17/3*6/11=34/11

a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)

=>(x+5)(x-3)+8=x^2-1

=>x^2+2x-15+8=x^2-1

=>2x-7=-1

=>x=3(loại)

b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)

=>(x-4)(x+1)+x^2+3+5(x-1)=0

=>x^2-3x-4+x^2+3+5x-5=0

=>2x^2+2x-6=0

=>x^2+x-3=0

=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)

e: =>x^2-2x+1+2x+2=5x+5

=>x^2+3=5x+5

=>x^2-5x-2=0

=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)

g: (x-3)(x+4)*x=0

=>x=0 hoặc x-3=0 hoặc x+4=0

=>x=0;x=3;x=-4

a) A = (x+3)² + (x-3)(x+3) - 2(x+2)(x - 4)

        = (x + 3)(x + 3) + (x - 3)(x + 3) - 2[x(x - 4) + 2(x - 4)]

        = x(x + 3) + 3(x + 3) + x(x + 3) - 3(x + 3) - 2[x² - 4x + 2x - 8]

        = x² + 3x + 3x + 9 + x² + 3x - 3x - 9 - 2(x² - 2x - 8)

        = x² + 3x + 3x + 9 +x² + 3x - 3x - 9 - 2x² + 4x + 16

        = (x² + x² - 2x²) + (3x + 3x + 3x - 3x + 4x) + (9 - 9 + 16) = 10x + 16

Thay x = -1/2 vào biểu thức trên ta có : \(10\cdot\left(-\frac{1}{2}\right)+16=-5+16=11\)

b) \(B=\left(3x+4\right)^2-\left(x-4\right)\left(x+4\right)-10x\)

\(B=9x^2+24x+16-x\left(x+4\right)+4\left(x+4\right)-10x\)

\(B=9x^2+24x+16-x^2-4x+4x+16-10x\)

\(B=\left(9x^2-x^2\right)+\left(24x-4x+4x-10x\right)+\left(16+16\right)\)

\(B=8x^2+14x+32\)

Thay x = -1/10 vào biểu thức trên ta có : \(B=8\cdot\left(-\frac{1}{10}\right)^2+14\cdot\left(-\frac{1}{10}\right)+32=\frac{767}{25}\)

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

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

\(C=x^2+2x+1-2x\left(2x-1\right)+1\left(2x-1\right)+3x^2-12\)

\(C=x^2+2x+1-4x^2+2x+2x-1+3x^2-12\)

\(C=\left(x^2-4x^2+3x^2\right)+\left(2x+2x+2x\right)+\left(1-1-12\right)\)

\(C=6x-12\)

Thay x = 1 vào biểu thức ta có : C = 6.1 - 12 = 6 -12 = -6

Còn bài kia làm nốt đi

1. Tính:

a)(x+4) (x^2 - 4x+16)=(x+4)(x²-4x+4²)=x³+64

b)(x-3y) (x^2 + 3xy+3y^2)=x³-(3y)³

c)(x-3)^2 -(x+3)^3=(x²-6x+9)-(x³+9x²+27x+27)=-8x²-33x-18-x³

d)(x^2-1)^3 - (x^2 +1)^3=(x²-1-x²-1)[(x²-1)²+(x²-1)(x²+1)+(x²+1)²]

=-2(x⁴-2x²+1+x⁴-1+x⁴+2x²+1)

=-2(3x⁴+2)=-6x⁴-4

a/ \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=0\)

\(\Leftrightarrow x^3-2x^2+4x+2x^2-4x+8-x^3-2x=0\)

\(\Leftrightarrow-2x=-8\)

\(\Leftrightarrow x=4\)

Vậy .....................

b/ \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow x^6-3x^4+3x^2-1-\left(x^6-x^4+x^4-x^2+x^2-1\right)=0\)

\(\Leftrightarrow x^6-3x^4+3x^2-1-x^6+1=0\)

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

\(\Leftrightarrow3x^2\left(-x^2+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x^2=0\\-x^2+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x^2=-1\Rightarrow x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\end{matrix}\right.\)

Vậy pt có 3 nghiệm là \(\left\{{}\begin{matrix}x=-1\\x=0\\x=1\end{matrix}\right.\)

| x | - 1 = 2

| x | = 2 + 1 

|x| = 3

X = 3 hoặc x = -3

chọn D. x = 3  hoặc x =  - 3

Bài 2:

a: a>b

=>2a>2b

=>2a+1>2b+1

mà 2b+1>2b-3

nên 2a+1>2b-3

b: 3x-1<=x+2

=>3x-x<=1+2

=>2x<=3

=>\(x\le\frac32\)

c: 2x+3>0

=>2x>-3

=>\(x>-\frac32\)

3x+1<x-4

=>3x-x<-4-1

=>2x<-5

=>\(x<-\frac52\)

2(x+1)+3>=3(5-x)

=>2x+2+3>=15-3x

=>2x+5>=15-3x

=>5x>=10

=>x>=2

\(\frac{x}{3}-\frac{x+1}{5}>1\)

=>\(\frac{5x-3\left(x+1\right)}{15}>1\)

=>5x-3(x+1)>15

=>5x-3x-3>15

=>2x>18

=>x>9

Bài 3:

Gọi độ dài quãng đường AB là x(km)

(Điều kiện: x>0)

Thời gian ô tô đi từ A đến B là \(\frac{x}{50}\) (giờ)

Thời gian ô tô đi từ B về A là \(\frac{x}{60}\) (giờ)

Tổng thời gian cả đi và về là 6h30p-1h=5h30p=5,5 giờ nên ta có:

\(\frac{x}{50}+\frac{x}{60}=5,5\)

=>\(\frac{6x}{300}+\frac{5x}{300}=5,5\)

=>\(x\cdot\frac{11}{300}=\frac{11}{2}\)

=>x=150(nhận)

Vậy: Độ dài quãng đường AB là 150km

Bài 1:

a: 2x-1=0

=>2x=1

=>\(x=\frac12\)

b: 3x-2=5+x

=>3x-x=2+5

=>2x=7

=>\(x=\frac72\)

c: 2(x-3)-4=3(1+x)-5x

=>2x-6-4=3+3x-5x

=>2x-10=3-2x

=>4x=13

=>\(x=\frac{13}{4}\)

d: \(\frac{x+1}{2}-\frac{2x}{3}=1\)

=>\(\frac{3\left(x+1\right)-4x}{6}=1\)

=>3(x+1)-4x=6

=>3x+3-4x=6

=>3-x=6

=>x=3-6=-3

e: x(x-2)+3(x-2)=0

=>(x-2)(x+3)=0

=>\(\left[\begin{array}{l}x-2=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-3\end{array}\right.\)

f: ĐKXĐ: x<>0; x<>1

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

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

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

=>4x=5

=>x=5/4(nhận)

Akai HarumaLovers

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