Lm giúp mik vs ạ
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\(M=\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\left|2-\sqrt{5}\right|}-\dfrac{2}{\left|2+\sqrt{5}\right|}\)
\(=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=\dfrac{8}{1}=8\)
Lm ơn giúp mik đii mà mik bt ơn bn đó nhiều lắm . Mik đang rất cần
1 interested in travelling by plane
2 15 minutes riding his bike to school
3 is made by my mother everymorning
4 take care of her little brother
5 able to speak 2 languages when he was young
6 able to play the piano
7 is played all over VN
8 to play voleyball very well when he was young
9 to go fishing when he was young
10 way to the shopping mall
over
camping
was
couldn't
percussion
takes
look after
go around
go through
wind
how long
for
do
Bài 17:
1) \(3^2-x^2=\left(3-x\right)\left(3+x\right)\)
2) \(x^2-36=\left(x-6\right)\left(x+6\right)\)
3) \(y^2-1=\left(y-1\right)\left(y+1\right)\)
4) \(25-y^2=\left(5-y\right)\left(5+y\right)\)
5) \(9x^2-1=\left(3x-1\right)\left(3x+1\right)\)
6) \(\dfrac{1}{25}-4x^2=\left(\dfrac{1}{5}-2x\right)\left(\dfrac{1}{5}+2x\right)\)
7) \(9x^2-y^2=\left(3x-y\right)\left(3x+y\right)\)
8) \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
Bài 18:
1) \(\left(x-5\right)\left(x+5\right)=x^2-25\)
2) \(\left(4-x\right)\left(4+x\right)=16-x^2\)
3) \(\left(x-\dfrac{2}{3}\right)\left(x+\dfrac{2}{3}\right)=x^2-\dfrac{4}{9}\)
4) \(\left(1+2x\right)\left(1-2x\right)=1-4x^2\)
5) \(-\left(2x+3\right)\left(3-2x\right)=\left(2x+3\right)\left(2x-3\right)=4x^2-9\)
6) \(-\left(5x-3\right)\left(3+5x\right)=\left(3-5x\right)\left(3+5x\right)=9-25x^2\)
7) \(-\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)=-\left(9x^2-\dfrac{4}{25}\right)=\dfrac{4}{25}-9x^2\)
8) \(-\left(2x-\dfrac{2}{3}\right)\left(2x+\dfrac{2}{3}\right)=-\left(4x^2-\dfrac{4}{9}\right)=\dfrac{4}{9}-4x^2\)
Bài 2:
a: Xét ΔABC có
X là trung điểm của BC
Y là trung điểm của AB
Do đó: XY là đường trung bình
=>XY//AC và XY=AC/2=3,5(cm)
hay XZ//AC và XZ=AC
b: Xét tứ giác AZBX có
Y là trung điểm của AB
Y là trung điểm của ZX
Do đó: AZBX là hình bình hành
mà \(\widehat{AXB}=90^0\)
nên AZBX là hình chữ nhật
d: Xét tứ giác AZXC có
XZ//AC
XZ=AC
Do đó: AZXC là hình bình hành
Bài 2
a)x-13=-22
x=-22+13
x=-9
b)(2x-8).5=2.53
(2x-8).5=2.125
(2x-8).5=250
2x-8=250:5
2x-8=50
2x=50+8
2x=58
⇒x=29
1 My house isn't far from the bank
2 The question were not as easy as I thought
3 It will take us 30 minutes to get there
4 She cooks very well
5 Hoa is shorter than Mai
hơi khó đọc nhỉ
Bài 1 :
1/\(y^3-y^4=y^3\left(1-y\right)\)
2/\(2x^4y^2-4x^2y^3+6x^2y^2=2x^2y^2\left(x^2-2y+3\right)\)
3/\(12x^2y-18xy^2-30y^2=6y\left(2x^2-3xy-5y\right)\)
4/\(3\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(3-y\right)\)
5/\(y\left(x-z\right)+7\left(z-x\right)=y\left(x-z\right)-7\left(x-z\right)=\left(x-z\right)\left(y-7\right)\)
6/\(27x^2\left(y-1\right)-9x^3\left(1-y\right)=27x^2\left(y-1\right)+9x^3\left(y-1\right)=9x^2\left(3+x\right)\)
7/\(5x\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(5x-x+y\right)=\left(x-y\right)\left(4x+y\right)\)
8/\(15x\left(x-1\right)-2\left(1-x\right)^2=15x\left(x-1\right)-2\left(x-1\right)^2=\left(x-1\right)\left(15x-2x+2\right)=\left(x-1\right)\left(13x+2\right)\)