Bài 3:

a: Ta có: \(4x\left(x+1\right)=8\left(x+1\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

b: Ta có: \(x\left(x-1\right)+2\left(x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

c: Ta có: \(2x\left(x-2\right)-\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-x+2\right)=0\)

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

2a) (3x-5y)(x+1)

b) (3x-2)(x-6)

c) (4y+1)(x-1)

d) (x-2)(x-3)(x-4)

e) (7x+1)(x-y)

d) pt<=> (x-2)(x-3)(x-4)=0

<=> x-2=0 hoặc x-3=0 hoặc x-4=0

<=> x=2 hoặc x=3 hoặc x=4

e) pt <=> (5x+1)(x-2)=0

<=> x=-1/5 hoặc x=2

g) pt <=> (5x-1)(x-2000)=0

<=> x=1/5 hoặc x=2000

h) pt <=> x(x-4)=0

<=> x =0 hoặc x=4

K) pt <=> x(x-1)=0

<=> x = 0 hoặc x=1

m) pt <=> x(6x+1)=0

<=> x = 0 hoặc x=-1/6

n) pt <=> (x+1)(1-x-1)=0

<=> -x(x+1)=0

<=> x = 0 hoặc x=-1

Bài 4:

\(a,\sqrt{64.\left(x-1\right)^2}=16\\ \Leftrightarrow8\sqrt{\left(x-1\right)^2}=16\\ \\ \Leftrightarrow\sqrt{\left(x-1\right)^2}=\dfrac{16}{8}=2\\ \left|x-1\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\\ b,\sqrt{4\left(x-2\right)}=8\\ \Leftrightarrow2\sqrt{x-2}=8\\ \Leftrightarrow\sqrt{x-2}=\dfrac{8}{2}=4\\ \Leftrightarrow x-2=4^2=16\\ \Leftrightarrow x=16+2=18\\ c,\dfrac{\sqrt{2x}}{\sqrt{8}}=5\\ \Leftrightarrow\dfrac{\sqrt{2}\sqrt{x}}{2\sqrt{2}}=5\\ \Leftrightarrow\dfrac{\sqrt{x}}{2}=5\\ \Leftrightarrow\sqrt{x}=5.2=10\\ \Leftrightarrow x=10^2=100\)

Bài 2:

\(a,\left(\sqrt{75}-2\sqrt{12}-\sqrt{27}\right).\sqrt{3}\\ =\left(\sqrt{3.5^2}-2.\sqrt{3.2^2}-\sqrt{3.3^2}\right).\sqrt{3}\\ =\left(5\sqrt{3}-2.2\sqrt{3}-3\sqrt{3}\right).\sqrt{3}\\ =-2\sqrt{3}.\sqrt{3}=-2.3=-6\\ b,\left(5\sqrt{2}-\sqrt{8}-\sqrt{98}\right):\sqrt{2}\\ =\left(5\sqrt{2}-\sqrt{2^2.2}-\sqrt{2.7^2}\right):\sqrt{2}\\ =\left(5\sqrt{2}-2\sqrt{2}-7\sqrt{2}\right):\sqrt{2}\\ =-4\sqrt{2}:\sqrt{2}=-4\)

\(c,\\ \dfrac{\sqrt{18}}{\sqrt{2}}=\sqrt{\dfrac{18}{2}}=\sqrt{9}=\sqrt{3^2}=3\\ d,\\ \sqrt{\dfrac{45}{7}}.\sqrt{\dfrac{28}{5}}=\sqrt{\dfrac{45.28}{7.5}}=\sqrt{\dfrac{9.5.4.7}{7.5}}=\sqrt{9.4}=\sqrt{36}=\sqrt{6^2}=6\)

3:

a: \(A=\sqrt{16\cdot a^2\cdot\left(a-1\right)^2}\)

=4|a(a-1)|

=4a(a-1)(Do a>=1)

=4a^2-4a

b: \(B=\dfrac{1}{a-5}\cdot7a^2\left|a-5\right|\)

\(=\dfrac{7a^2\cdot\left(a-5\right)}{a-5}\left(a>5\right)\)

=7a^2

c: \(=\sqrt{225a^2}-3a=15a-3a\left(a>=0\right)\)

=12a

d:D \(=\left(a-3\right)^2-\sqrt{36a^2}\)

=(a-3)^2-6|a|

TH1: a>=0

D=(a-3)^2-6a=a^2-12a+9

TH2: a<0

D=(a-3)^2+6a=a^2+9

e: \(C=\sqrt{225a^2}-8a=15a-8a=7a\)

Tham khảo:

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giúp mình với mình bế tắc r =(

Gián đoạn

Gián đoạn

gián đoạn

Câu 2)

b, Công thực hiện là

\(A=P.h=10m.h=10.6.12=720J\) 

a, Phụ thuộc vào khối lượng và quãng đường vật di chuyển

A = F.s

A : công cần tính ( đơn vị J(Jun) ; 1kJ (ki lô jun)= 1000J)

F : lực vật tác động

s : quãng đường vật di chuyển

Câu 3)

Để tính công thực hiện trong 1s

P = A/t

P : công suất ( đơn vị W (oát) -1MGW = 1000kW = 1,000,000W)

A : công

t : thời gian (đơn vị giây)

b, Công suất thực hiện là

\(P=\dfrac{A}{t}=\dfrac{P.h}{t}=\dfrac{10m.h}{t}\\ =\dfrac{10.50.8}{50}=80W\)

Câu 4)

5p = 300s

10p = 600s

Công suất leo của Hương là

\(P=\dfrac{A}{t}=\dfrac{P.h}{t}\\ =\dfrac{10m.h}{t}=\dfrac{10.40.6}{300}=8W\)

Công suất leo của Lan là

\(P=\dfrac{10m.h}{t}=\dfrac{10.50.9}{600}=7,5W\)

Hương leo khoẻ hơn do 8 W > 7,5 W