Bài 4:

a) \(\dfrac{-2x+4}{x^2-4}=\dfrac{?}{x+2}\)

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

b) \(\dfrac{x^2+3x}{3x+9}=\dfrac{?}{3}\)

\(\Rightarrow?=\dfrac{\left(x^2+3x\right)\cdot3}{3x+9}=\dfrac{3x\left(x+3\right)}{3\left(x+3\right)}=\dfrac{3x}{3}=x\)

c) \(\dfrac{x^2-1}{x-1}=\dfrac{?}{x+1}\)

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

d) \(\dfrac{x^2-5x+6}{x-3}=\dfrac{x-2}{?}\)

\(\Rightarrow?=\dfrac{\left(x-3\right)\left(x-2\right)}{x^2-5x+6}=\dfrac{\left(x-3\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=1\)

mn giúp em vs ạ 

Bài 1:

a) Ta có:

\(VT=\dfrac{3x+6}{x+2}=3\) (ĐK: \(x\ne-2\))

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

\(=\dfrac{3}{1}\)

\(=3=VP\left(dpcm\right)\)

b) Ta có:

\(VT=\dfrac{x^2+2x}{3x+6}\) (ĐK: \(x\ne-2\))

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

\(=\dfrac{x}{3}=VP\left(dpcm\right)\)

c) Ta có:

\(VT=\dfrac{x-1}{x^2-1}\) (ĐK: \(x\ne\pm1\))

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

\(=\dfrac{1}{x+1}=VP\left(dpcm\right)\)

d) Ta có:

\(VT=\dfrac{x^2+3x-4}{x-1}\) (ĐK: \(x\ne1\))

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

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

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

\(=x+4=VP\left(dpcm\right)\)

Bài 2:

Với \(x\ne0\)

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

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

Vậy cả 3 phân thức đều bằng nhau.

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

\(A=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+2\left(x^2-9\right)\)

\(A=3x^2-6x+3-x^2-2x-1+2x^2-18\)

\(A=4x^2-8x--16\)

\(B=5x\left(x-7\right)\left(x+7\right)-2\left(2x-1\right)^2\)

\(B=5x\left(x^2-49\right)-2\left(4x^2-4x+1\right)\)

\(B=5x^3-245x-8x^2+8x-2\)

\(B=5x^3-8x^2-253x-2\)

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

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

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

\(C=x^4-x^2\)

\(D=\left(4x-1\right)\left(1+16x^2+4x\right)-\left(4x+1\right)^3\)

\(D=\left(4x-1\right)\left(16x^2+4x+1\right)-\left(64x^3+48x^2+12x+1\right)\)

\(D=64x^3-1-64x^3-48x^2-12x-1\)

\(D=-48x^2-12x-2\)

\(A=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+2\left(x^2-3^2\right)\\ =3x^2-6x+3-x^2-2x-1+2x^2-18\\ =4x^2-8x-16\)

\(B=5x\left(x^2-7^2\right)-2\left(4x^2-4x+1\right)\\ =5x^3-245x-8x^2+8x-2\\ =5x^3-8x^2-237x-2\)

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

\(D=\left(4x\right)^3-1^3-\left(64x^3+48x^2+12x+1\right)\\ =64x^3-1-64x^3-48x^2-12x-1\\ =-48x^2-12x-2\)

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

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

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

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

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

(x−1)2(x+1)

x² - 4x - y² + 4

= (x² - 4x + 4) - y²

= (x - 2)² - y²

= (x - y - 2)(x + y - 2)

(x - y - 2)(x + y - 2)

Bài 3:

\(I=64^2-72\cdot64+36^2\)

\(I=64^2-2\cdot36\cdot64+36^2\)

\(I=\left(64-36\right)^2\)

\(I=28^2\)

\(I=784\)

\(K=47\cdot53\)

\(K=\left(50-3\right)\left(50+3\right)\)

\(K=50^2-3^2\)

\(K=2500-9\)

\(K=2491\)

\(M=123^3-69\cdot123^2+369\cdot23^2-23^3\)

\(M=123^3-3\cdot23\cdot123^2+3\cdot123\cdot23^2-23^3\)

\(M=\left(123-23\right)^3\)

\(M=100^3\)

\(M=1000000\)

\(N=54^3+138\cdot54^2+162\cdot46^2+46^3\)

\(N=54^3+3\cdot46\cdot54^2+3\cdot54\cdot46^2+46^3\)

\(N=\left(54+46\right)^3\)

\(N=100^3\)

\(N=1000000\)

Bài 1

A = 3(x - 1)² - (x + 1)² + 2(x - 3)(x + 3)

= 3x² - 6x + 1 - x² - 2x - 1 + 2x² - 18

= (3x² - x² + 2x²) + (-6x - 2x) + (1 - 1 - 18)

= 4x² - 8x - 18

B = 5x(x - 7)(x + 7) - 2(2x - 1)²

= 5x³ - 245x - 8x² + 8x - 2

= 5x³ - 8x² + (-245x + 8x) - 2

= 5x³ - 8x² - 237x - 2

C = (x² - 2)(x² + 2) - x² + 4

= x⁴ - 4 - x² + 4

= x⁴ - x²

D = (4x - 1)(1 + 16x + 4x) - (4x + 1)³

= 4x + 80x² - 1 - 20x - 64x³ - 48x² - 12x - 1

= -64x³ + (80x² - 48x²) + (4x - 20x - 12x) + (-1 - 1)

= -64x³ + 32x² - 28x - 2

\(=x\left(a-b\right)-\left(a^2-2ab+b^2\right)=\)

\(=\left(a-b\right)x-\left(a-b\right)^2=\)

\(=\left(a-b\right)\left[x-\left(a-b\right)\right]=\left(a-b\right)\left(x-a+b\right)\)

\(x^2-2018x=0\\ \Leftrightarrow x\left(x-2018\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2108=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2018\end{matrix}\right.\)

Vậy `x=0` hoặc `x=2018`

\(2x^2+5x=0\\ \Leftrightarrow x\left(2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)

Vậy `x=0` hoặc `x=-5/2`

Bài 2:

\(E=x^2-4x+6=x^2-4x+4+2\\ =\left(x-2\right)^2+2\ge2\)

GTNN của E đạt `2` khi và chỉ khi `x=2`

\(F=25x^2+20x+8=\left(5x\right)^2+2.5x.2+4+4\\ =\left(5x+2\right)^2+4\ge4\)

GTNN của F đạt `4` khi và chỉ khi `x=-2/5`

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

GTNN của G đạt \(-\dfrac{1}{4}\) khi và chỉ khi `x=1/2`

\(H=-4x^2-4x-3=-\left(4x^2+4x+3\right)\\ =-\left(\left(2x\right)^2+2.2x.1+1+2\right)\\ =-\left(2x+1\right)^2-2\le-2\)

GTLN của H đạt `-2` khi và chỉ khi `x=-1/2`

$HaNa$☘

cảm ơn bn nha

a: Xét ΔABC vuông tại  A và ΔHBA vuông tại H có

góc B chung

=>ΔABC đồng dạng với ΔHBA

b: Xét tứ giác ADHE có

góc ADH=góc AEH=góc DAE=90 độ

=>ADHE là hình chữ nhật

=>AH=DE

ΔABC vuông tại A có AH là đườg cao

nên BA^2=BH*BC; AH^2=HB*HC

=>DE^2=4*9=36 và AB^2=4*13=52

=>AB=2*căn 13(cm); DE=6cm

c: ΔAHB vuông tại H có HD là đường cao

nên AD*AB=AH^2

ΔAHC vuông tại H có HE là đường cao

nên AE*AC=AH^2

=>AD*AB=AE*AC