a) ĐK: \(x>1\)

\(P=\dfrac{2}{\left(\sqrt{x+1}\right)^3+\left(\sqrt{x-1}\right)^3}\cdot\dfrac{\dfrac{2x}{\sqrt{x-1}}-\dfrac{\sqrt{\left(x+1\right)\left(x-1\right)}}{\sqrt{x-1}}}{\dfrac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x-1}\cdot\sqrt{x+1}}}\)

\(P=\dfrac{2}{\left(\sqrt{x+1}+\sqrt{x-1}\right)\left(x+1-\sqrt{\left(x+1\right)\left(x-1\right)}+x-1\right)}\cdot\dfrac{2x-\sqrt{\left(x+1\right)\left(x-1\right)}}{\sqrt{x-1}}\cdot\dfrac{\sqrt{x-1}\cdot\sqrt{x+1}}{\sqrt{x+1}-\sqrt{x-1}}\)

\(P=\dfrac{2}{\left(\sqrt{x+1}+\sqrt{x-1}\right)\left(2x-\sqrt{\left(x+1\right)\left(x-1\right)}\right)}\cdot\dfrac{2x-\sqrt{\left(x+1\right)\left(x-1\right)}}{\sqrt{x-1}}\cdot\dfrac{\sqrt{x-1}\cdot\sqrt{x+1}}{\sqrt{x+1}-\sqrt{x-1}}\)

\(P=\dfrac{2}{\left(\sqrt{x+1}+\sqrt{x-1}\right)}\cdot\dfrac{\sqrt{x+1}}{\left(\sqrt{x+1}-\sqrt{x-1}\right)}\)

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

\(P=\dfrac{2\sqrt{x+1}}{x+1-x+1}\)

\(P=\dfrac{2\sqrt{x+1}}{2}=\sqrt{x+1}\)

b)

\(P=x-1\Leftrightarrow\sqrt{x+1}=x-1\)

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

\(\Leftrightarrow x+1=x^2-2x+1\)

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

\(\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=3\end{matrix}\right.\)

Vậy x = 3 là GT cần tìm

\(2\left(x+2\right)^2-x^3-8=0\)

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

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

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

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

\(\Leftrightarrow-x\left(x+2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=4\end{matrix}\right.\)

Vậy \(S=\left\{0;-2;4\right\}\)

\(\left\{{}\begin{matrix}\dfrac{x-5y}{4}-\dfrac{3x-2y}{3}=-\dfrac{25}{24}\\\dfrac{5x+3y}{8}+\dfrac{x-6y}{5}=\dfrac{33}{80}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3\left(x-5y\right)-4\left(3x-2y\right)}{12}=-\dfrac{25}{24}\\\dfrac{5\left(5x+3y\right)+8\left(x-6y\right)}{40}=\dfrac{33}{80}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x-15y-12x+8y}{12}=-\dfrac{25}{24}\\\dfrac{25x+15y+8x-48y}{40}=\dfrac{33}{80}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2\left(-9x-7y\right)}{24}=-\dfrac{25}{24}\\\dfrac{2\left(33x-33y\right)}{80}=\dfrac{33}{80}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}18x+14y=25\\2x-2y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}18x+14y=25\\14x-14y=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}32x=32\\2x-2y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\2\cdot1-2y=1\end{matrix}\right.\)

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

Gọi chữ số hàng chục của số cần tìm là x (đv) (ĐK : \(x\in N,0< x< 10\))

Gọi chữ số hàng đơn vị của số cần tìm là y (đv) (ĐK:\(y\in N,7< y< 10\) )

Vì chữ số hàng đơn vị hơn chữ số hàng chục 7 đơn vị nên ta có PT(1):

y - x = 7 \(\Leftrightarrow\) -x + y = 7

 Số cần tìm có dạng: 10x + y

Nếu đổi chỗ hai chữ số cho nhau và viết thêm chữ số 0 vào giữa hai chữ số ấy thì số mới có dạng: 100y + x

Vì số mới lớn hơn 3 lần số ban đầu là 815 đơn vị nên ta có PT(2):

\(\left(100y+x\right)-3\left(10x+y\right)=815\)

\(\Leftrightarrow100y+x-30x-3y=815\Leftrightarrow-29x+97y=815\)

Từ (1)(2) ta có HPT:

\(\left\{{}\begin{matrix}-x+y=7\\-29x+97y=815\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-29x+29y=203\\-29x+97y=815\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-68y=-612\\-x+y=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=9\\-x+9=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=9\\x=2\end{matrix}\right.\)(T/m)

Vậy số cần tìm là 29

1.plant

2. is sleeping

3. stayed, watched

4. are doing

5. are traveling

6. were watching

7. will take over

8. are learning

9. have they lived

10. Do your neighbors love

a) \(\dfrac{2^{12}\cdot5^7+4^6\cdot25^3}{8^5\cdot25^3+\left(2^2\cdot5^6\right)}\)

\(=\dfrac{2^{12}\cdot5^6\cdot5+\left(2^2\right)^6\cdot\left(5^2\right)^3}{\left(2^3\right)^5\cdot\left(5^2\right)^3+\left(2^2\right)^6\cdot5^6}\)

\(=\dfrac{2^{12}\cdot5^6\cdot5+2^{12}\cdot5^6}{2^3\cdot2^{12}\cdot5^6+2^{12}\cdot5^6}\)

\(=\dfrac{2^{12}\cdot5^6\cdot\left(5+1\right)}{2^{12}\cdot5^6\left(2^3+1\right)}\)

\(=\dfrac{6}{8+1}=\dfrac{6}{9}=\dfrac{2}{3}\)

b) \(\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)

\(=\dfrac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot\left(3^2\right)^2}{\left(2^2\right)^6\cdot3^6+\left(2^3\right)^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-\left(5^2\right)^5\cdot\left(7^2\right)^2}{\left(5^3\right)^3\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{2^{12}\cdot3^4\cdot3-2^{12}\cdot3^4}{2^{12}\cdot3^4\cdot3^2+2^{12}\cdot3^4\cdot3}-\dfrac{5^9\cdot7^3\cdot5-5^9\cdot7^3\cdot5\cdot7}{5^9\cdot7^3+5^9\cdot7^3\cdot8}\)

\(=\dfrac{2^{12}\cdot3^4\cdot\left(3-1\right)}{2^{12}\cdot3^4\cdot\left(3^2+3\right)}-\dfrac{5^9\cdot7^3\cdot\left(5-35\right)}{5^9\cdot7^3\cdot\left(1+8\right)}\)

\(=\dfrac{3-1}{9+3}-\dfrac{5-35}{1+8}\)

\(=\dfrac{1}{6}-\dfrac{-30}{9}=\dfrac{7}{2}\)

\(\left\{{}\begin{matrix}x+3y=-2\\5x-4y=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\x+3y=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\x+3\cdot\dfrac{-21}{19}=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\x=\dfrac{25}{19}\end{matrix}\right.\)

Vậy HPT có nghiệm duy nhất (x;y) = \(\left(\dfrac{25}{19};-\dfrac{21}{19}\right)\)

Go: cycling, skating, swimming, camping

Do: athletics, karate, judo, aerobics, gardening, gymnastics

Collect: bottles, stamps, dolls, coins

Play: basketball, football, tennis, computer games, table tennis

Take: photos

Watch: TV, films, cartoons,music videos

1. C

2.A

3. C

4. D

5. B

6. A

7. D

8.B

\(\left(x-3\right)^2-2\left(x-3\right)=4x-14\)

\(\Leftrightarrow x^2-6x+9-2x+6=4x-14\)

\(\Leftrightarrow x^2-12x+29=0\)

\(\Delta'=\left(-6\right)^2-1\cdot29=36-29=7>0\)

\(\rightarrow\) PT có 2 nghiệm pb

\(x_1=\dfrac{6+\sqrt{7}}{1}=6+\sqrt{7}\)

\(x_2=\dfrac{6-\sqrt{7}}{1}=6-\sqrt{7}\)

Vậy S = \(\left\{6+\sqrt{7};6-\sqrt{7}\right\}\)