Số xe cửa hàng nhập về là:

\(80:32\%=250\) (chiếc)

Số xe cửa hàng còn lại là:
`250 - 80 =170` (chiếc) 

ĐS: ...

\(a)2^3-50:25+13\cdot7=8-2+91\\ =6+91\\ =97\\ b)60-\left[120-\left(42-33\right)\cdot2\right]\\ =60-\left(120-9\cdot2\right)\\ =60-\left(120-18\right)\\ =60-102\\ =-42\\ c)3^{17}:3^{15}+8\cdot3\\ =3^{17-15}+24\\ =3^2+24\\ =9+24\\ =33\\ d)12:\left\{390:\left[500-\left(125+35\cdot7\right)\right]\right\}\\ =12:\left\{390:\left[500-\left(125+245\right)\right]\right\}\\ =12:\left[390:\left(500-370\right)\right]\\ =12:\left(390:130\right)\\ =12:3=4\)

e: \(72^3\cdot49-72^2\cdot9\)

\(=72^2\left(72\cdot49-9\right)\)

\(=5184\cdot3519=18242496\)

f: \(\dfrac{2^3+2^4+2^5}{7^2}=\dfrac{2^3\left(1+2+2^2\right)}{7^2}=\dfrac{8\cdot7}{49}=\dfrac{8}{7}\)

g: \(\dfrac{15^{22}\cdot7^{18}}{7^{20}\cdot15^{21}}=\dfrac{15^{22}}{15^{21}}\cdot\dfrac{7^{18}}{7^{20}}=\dfrac{15}{7^2}=\dfrac{15}{49}\)

Diện tích của các hình là: 

\(\left(2+2\right)\times\left(3+3\right)=24\left(dm^2\right)\)

Diện tích phần màu trắng là:

\(\dfrac{1}{2}\times2\times3+\dfrac{1}{2}\times2\times3+\dfrac{1}{2}\times2\times3+\dfrac{1}{2}\times2\times3=12\left(dm^2\right)\)

Diện tích phần tô màu là:

\(24-12=12\left(dm^2\right)\)

ĐS: ...

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{2}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{2}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}-\dfrac{2}{y}=2\\\dfrac{3}{x}+\dfrac{2}{y}=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{4}{x}-\dfrac{2}{y}+\dfrac{3}{x}+\dfrac{2}{y}=2+5=7\\\dfrac{3}{x}+\dfrac{2}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{x}=7\\\dfrac{2}{y}=5-\dfrac{3}{x}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1\\\dfrac{2}{y}=5-\dfrac{3}{1}=5-3=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

=>Chọn C

a: ta có: \(\widehat{MNS}=\widehat{HNQ}\)(hai góc đối đỉnh)

mà \(\widehat{HNQ}=60^0\)

nên \(\widehat{MNS}=60^0\)

b: Ta có: \(\widehat{QNH}=\widehat{PMN}\left(=60^0\right)\)

mà hai góc này là hai góc ở vị trí đồng vị

nên PI//QS

=>MP//NQ

c: ta có: MP//NQ

KP\(\perp\)MP

Do đó: KP\(\perp\)QN

d: ta có: MI//SN

=>\(\widehat{MIS}+\widehat{S}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{S}+100^0=180^0\)

=>\(\widehat{S}=80^0\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\y>=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}2\sqrt{x}+\sqrt{y}=5\\3\sqrt{x}-\sqrt{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}+\sqrt{y}+3\sqrt{x}-\sqrt{y}=5+1\\2\sqrt{x}+\sqrt{y}=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5\sqrt{x}=6\\\sqrt{y}=5-2\sqrt{x}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=\dfrac{6}{5}\\\sqrt{y}=5-2\cdot\dfrac{6}{5}=5-\dfrac{12}{5}=\dfrac{13}{5}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{36}{25}\\y=\dfrac{169}{25}\end{matrix}\right.\)

=>Chọn B

\(\left\{{}\begin{matrix}2\sqrt{x}+\sqrt{y}=5\\3\sqrt{x}-\sqrt{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{x}=6\\2\sqrt{x}+\sqrt{y}=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=\dfrac{6}{5}\\\dfrac{12}{5}+\sqrt{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{36}{25}\\\sqrt{y}=5-\dfrac{12}{5}=\dfrac{13}{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{36}{25}\\y=\left(\dfrac{13}{5}\right)^2=\dfrac{169}{25}\end{matrix}\right.\)

=> Chọn B

a) Ta có: 

\(VT=\left(a+b\right)^2-4ab=\left(a^2+2ab+b^2\right)-4ab\\ =a^2+2ab+b^2-4ab=a^2-2ab+b^2\\ =\left(a-b\right)^2=VP\)

=> Đpcm 

b) Ta có:

\(VT=\left(a-b\right)^3=\left[-\left(b-a\right)\right]^3=\left[\left(-1\right)\cdot\left(b-a\right)\right]^3\\ =\left(-1\right)^3\left(b-a\right)^3=\left(-1\right)\cdot\left(b-a\right)^3=-\left(b-a\right)^3=VP\)

=> Đpcm  

c) Ta có: 

\(\left(n+2\right)^2-n^2=\left(n^2+4n+4\right)-n^2\\ =n^2+4n+4-n^2=4n+4=4\left(n+1\right)⋮4\forall n\in N\) 

=> Đpcm 

a: \(\left(a+b\right)^2-4ab\)

\(=a^2+2ab+b^2-4ab\)

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

b: \(\left(a-b\right)^3=\left[-\left(b-a\right)\right]^3=-\left(b-a\right)^3\)

c: \(\left(n+2\right)^2-n^2=\left(n+2+n\right)\left(n+2-n\right)\)

\(=2\left(2n+2\right)=4\left(n+1\right)⋮4\)

e: \(2^{5x-4}=64\)

=>\(2^{5x-4}=2^6\)

=>5x-4=6

=>5x=10

=>x=10/5=2

f: \(2^{3x+2}=4^{x+6}\)

=>\(2^{3x+2}=2^{2x+12}\)

=>3x+2=2x+12

=>3x-2x=12-2

=>x=10

g: \(4^x=5\cdot4^3-4\cdot4^3\)

=>\(4^x=4^3\)

=>x=3

h: \(4^{5x-3}=16^{2x-1}\)

=>\(4^{5x-3}=\left(4^2\right)^{2x-1}=4^{4x-2}\)

=>5x-3=4x-2

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

=>x=1

i: \(5^{7x-2}=5^{3x+10}\)

=>7x-2=3x+10

=>4x=12

=>x=4

l: \(\dfrac{16}{2^x}=2\)

=>\(2^x=\dfrac{16}{2}=8=2^3\)

=>x=3

m: \(\dfrac{\left(-3\right)^x}{81}=-27\)

=>\(\left(-3\right)^x=\left(-3\right)^3\cdot\left(-3\right)^4=\left(-3\right)^7\)

=>x=7

\(e)2^{5x-4}=64\\ \Rightarrow2^{5x-4}=2^6\\ \Rightarrow5x-4=6\\ \Rightarrow5x=6+4=10\\ \Rightarrow x=\dfrac{10}{5}\\ \Rightarrow x=2\\ f)2^{3x+2}=4^{x+6}\\ \Rightarrow2^{3x+2}=\left(2^2\right)^{x+2}\\ \Rightarrow2^{3x+2}=2^{2x+4}\\ \Rightarrow3x+2=2x+4\\ \Rightarrow3x-2x=4-2\\ \Rightarrow x=2\\ g)4^x=5\cdot4^3-4\cdot4^3\\ \Rightarrow4^x=4^3\cdot\left(5-4\right)\\ \Rightarrow4^x=4^3\\ \Rightarrow x=3\\ h)4^{5x-3}=16^{2x-1}\\ \Rightarrow4^{5x-3}=\left(4^2\right)^{2x-2}\\ \Rightarrow4^{5x-3}=4^{4x-4}\\ \Rightarrow5x-3=4x-4\\ \Rightarrow5x-4x=-4+3\\ \Rightarrow x=-1\\ i)5^{7x-2}=5^{3x+10}\\ \Rightarrow7x-2=3x+10\\ \Rightarrow7x-3x=10+2\\ \Rightarrow4x=12\\ \Rightarrow x=12:4\\ \Rightarrow x=3\)