Giúp mình bài này vs nha mình đag cần gấp lắm . Mình cảm ơn nhiều lắm ><
Talk about a traditional craft village you knows (khoảng 5-6 câu )
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1) Vì x=25 thỏa mãn ĐKXĐ nên Thay x=25 vào biểu thức \(A=\dfrac{\sqrt{x}-2}{x+1}\), ta được:
\(A=\dfrac{\sqrt{25}-2}{25+1}=\dfrac{5-2}{25+1}=\dfrac{3}{26}\)
Vậy: Khi x=25 thì \(A=\dfrac{3}{26}\)
2) Ta có: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}+\dfrac{2x+8\sqrt{x}-6}{x-\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-5\sqrt{x}+6+2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3x+3\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
\(a,A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)
\(=x-0,2-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)
\(=\left(-0,2-2+2\right)+\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)\)
\(=-0,2\)
\(b,B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)
\(=x^3-8y^3-x^3+8y^3-10\)
\(=-10\)
\(c,C=4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)-4x\)
\(=4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)
\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)
\(=13\)
a) \(A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)
\(A=x-\dfrac{1}{5}-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)
\(A=\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)-\left(\dfrac{1}{5}+2-2\right)\)
\(A=-\dfrac{1}{5}\)
Vậy: ...
b) \(B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)
\(B=\left[x^3-\left(2y\right)^3\right]-\left[x^3-\left(2y\right)^3\right]-10\)
\(B=-10\)
Vậy: ...
c) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x+1\right)\left(x-1\right)-4x\)
\(=4\left(x^2+2x+4\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)
\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)
\(=\left(4x^2+4x^2-8x^2\right)+\left(8x-4x-4x\right)+\left(4+1+8\right)\)
\(=13\)
Vậy:...
Refer
1. “Your cousin speaks English very well” Paul told me
Paul said that ___________my cousin spoke English very well____________
2. “The man broke out of prison yesterday” said the policeman
The policeman told us_that the man had broken out of prison the day beforr__
3. “I’ll lend you this book as soon as I finish it” Owen said to me
Owen said __me that he would lend me that book as soon as he finished it___
4. “I think I forgot to turn off the lights this morning” Brenda told Brian
Brenda told Brian ____that he thought he had forgotten to turn off the lights that morning.____
5. “I work eight hours a day, except when the children are on holiday” said Mrs. Wood
Mrs. Wood said me that he worked eight hours a day, excepted when the children were on holiday
6. “You’ve been making good progress this semester” Miss Lynn told me
Miss Lynn said that _____I had been making good progress that semester_________
7. “If you bought all the tickets, you would win the lottery” the man said
The man told me ______that If I had bought all the tickets, I would win the lottery______________
8. “I like swimming but I don’t go very often” Jill said to Pam
Jill said that ______he liked swimming but he didn’t go very often___________________________
9. “I want to buy it, but I haven’t brought any money” said Patrick
Patrick told me _________that he wanted to buy it, but he hadn’t brought any money_______________________
10. “I’m going to visit my aunt in Hue, but I’m not sure when” said Mai
Mai told me _________that she was going to visit her aunt in Hue, but she was not sure when__________________
1) Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:
\(BC^2=AB^2+AC^2\)
\(\Leftrightarrow BC^2=6^2+8^2=100\)
hay BC=10(cm)
Áp dụng hệ thức lượng trong tam giác vuông vào ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC, ta được:
\(AH\cdot BC=AB\cdot AC\)
\(\Leftrightarrow AH\cdot10=6\cdot8=48\)
hay AH=4,8(cm)
Theo định lí \(sin\):
\(\dfrac{sin\alpha}{F_1}=\dfrac{sin\beta}{F_2}=\dfrac{sin\gamma}{F}\)\(\Rightarrow F_2=\dfrac{F_1}{sin\alpha}\cdot sin\beta\)
\(F_{min}\Leftrightarrow sin\alpha=1\Rightarrow\alpha=90^o\)
\(\Rightarrow\beta=120-90=60^o\)
\(\Rightarrow F_2=\dfrac{6}{1}\cdot sin60^o=3\sqrt{3}N\)
Coi điểm tựa G là trung điểm AB.
Ta có hệ:
\(\left\{{}\begin{matrix}AH+HB=L=80\\3AH=HB\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}AH=20\\BH=60\end{matrix}\right.\)
\(\Rightarrow HG=\dfrac{1}{4}AB=\dfrac{1}{4}\cdot80=20cm\)
\(\Rightarrow GB=BH-HG=60-20=40cm\)