1 At this time they are writing their homework

2 Are you having breakfast at the moment?

3 Is your mother cooking in the kitchen

4 Mr Peter isn't listening to music right now

5 Lan and I are playing tennis at present

Đề 1:

Bài 1:

\(a,=\sqrt{\left(\sqrt{7}+1\right)^2}-\left|-1+\sqrt{7}\right|=\sqrt{7}+1-\sqrt{7}+1=2\\ b,=2\sqrt{2}-4\sqrt{2}-5\sqrt{2}+\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2}-7\sqrt{2}=\dfrac{-13\sqrt{2}}{\sqrt{2}}\)

Bài 2:

\(PT\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=\dfrac{1}{2}\Leftrightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\dfrac{1}{2}=1\\x=-\dfrac{1}{2}+\dfrac{1}{2}=0\end{matrix}\right.\)

Bài 3:

\(a,M=\dfrac{a-2\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{2\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)}=\dfrac{2}{\sqrt{a}+1}\\ b,M< 1\Leftrightarrow\dfrac{2}{\sqrt{a}+1}-1< 0\Leftrightarrow\dfrac{1-\sqrt{a}}{\sqrt{a}+1}< 0\\ \Leftrightarrow1-\sqrt{a}< 0\left(\sqrt{a}+1>0\right)\\ \Leftrightarrow a>1\)

Câu 1:

\(\left(4x+3\right)\left(3x^2+x-2\right)\left(2x^2-3x-5\right)=0\\ \Leftrightarrow\left(4x+3\right)\left(3x-2\right)\left(x+1\right)\left(2x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-1\\x=\dfrac{2}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow A=\left\{-1;-\dfrac{3}{4};\dfrac{2}{3};\dfrac{5}{2}\right\}\)

Câu 2:

\(\left(x^2-4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\Leftrightarrow A=\left\{-2;2;3\right\}\\ \left|5x\right|-11\le0\Leftrightarrow\left|5x\right|\le11\Leftrightarrow-11\le5x\le11\\ \Leftrightarrow-\dfrac{11}{5}\le x\le\dfrac{11}{5}\\ \Leftrightarrow B=\left[-\dfrac{11}{5};\dfrac{11}{5}\right]\)

\(\Leftrightarrow A\cap B=\left\{-2;2\right\}\\ A\cup B=\left[-\dfrac{11}{5};3\right]\\ A\B=\left\{3\right\}\)

Bài 4:

\(P=\dfrac{4x^2-2x+7}{2x-1}=\dfrac{2x\left(2x-1\right)+7}{2x-1}=2x+\dfrac{7}{2x-1}\in Z\\ \Leftrightarrow2x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-3;0;1;4\right\}\\ Q=\dfrac{4x^2-2x+3}{2x-1}=\dfrac{2x\left(2x-1\right)+3}{2x-1}=2x+\dfrac{3}{2x-1}\in Z\\ \Leftrightarrow2x-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-1;0;1;2\right\}\)

Bài 5:

\(M=\dfrac{\left(5x-1\right)\left(5x+1\right)}{1-5x}+\dfrac{\left(y-3\right)\left(5x+1\right)}{y-3}=-\left(5x+1\right)+5x+1=0\)

Bài 6:

\(VT=\dfrac{a\left(a+3b\right)}{\left(a+3b\right)\left(a-3b\right)}-\dfrac{\left(2a+b\right)\left(a-3b\right)}{\left(a-3b\right)^2}=\dfrac{a}{a-3b}-\dfrac{2a+b}{a-3b}=\dfrac{-a-b}{a-3b}\)

\(VP=\dfrac{\left(a+b\right)\left(a+c\right)}{\left(a+c\right)\left(3b-a\right)}=\dfrac{a+b}{3b-a}=\dfrac{-a-b}{a-3b}\)

Vậy ta đc đpcm

ĐKXĐ: x>=0; x<>9

\(B=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

one - by- of - like - saying - no - made - about - who - never

Câu 2.

\(C=\dfrac{N}{20}=\dfrac{4080}{20}=204\)(chu kì)

\(L=\dfrac{N}{2}\cdot3,4=\dfrac{4080}{2}\cdot3,4=6936A^o\)

\(G=X=30\%\cdot4080=1224nu\)

\(A=T=\dfrac{4080-2\cdot1224}{2}=816nu\)

\(a.C=\dfrac{N}{20}=204\)

\(b.L=\dfrac{Nx3,4}{2}=6936A\)

\(c.A=T=20\%N=816;G=X=30\%N=1224\)

\(d.rN=\dfrac{N}{2}=2040\)

b.L=Nx3,4/2=6936A