Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
I.Choose the word having the underlined letters pronouced differently from the others.
1. A. game B. geography C. vegetable D. change ( mình đánh trọng âm chữ '' g'') có gì sai thì bạn sửa nhad
2. A. watches B. brushes C. classes D. lives
3. A. their B. math C.thing D. theater
4. A. read B. teacher C. eat D. ahead
5. A. like B. light C. twice D. lips
6. A. never B. tennis C. soccer D. tent
7. A. meat B. tea C. weather D. bean
8. A. movie B. jog C. bottle D. orange
Ta có: \(\frac{2011113}{2011114}=\frac{2011114-1}{2011114}=1-\frac{1}{2011114}\)
\(\frac{2011114}{2011115}=\frac{2011115-1}{2011115}=1-\frac{1}{2011115}\)
Vì \(\frac{1}{2011114}>\frac{1}{2011115}\) nên \(1-\frac{1}{2011114}< 1-\frac{1}{2011115}\)
Suy ra: \(\frac{2011113}{2011114}< \frac{2011114}{2011115}\)
a)
\(1+tan^2a=\frac{1}{cos^2a}\)
\(1+3^2=\frac{1}{cos^2a}\)
\(10=\frac{1}{cos^2a}\)
\(cos^2a=\frac{1}{10}\)
\(cosa=\pm\sqrt{\frac{1}{10}}=\pm\frac{1}{\sqrt{10}}\)
\(sin^2a+cos^2a=1\)
\(sin^2a+\frac{1}{10}=1\)
\(sin^2a=\frac{9}{10}\)
\(sina=\pm\sqrt{\frac{9}{10}}=\pm\frac{3}{\sqrt{10}}\)
Vì tan = 3 nên M có 2 trường hợp :
TH1 :
sin và cos cùng dương
\(\Rightarrow M=\frac{\frac{1}{\sqrt{10}}+\frac{3}{\sqrt{10}}}{\frac{1}{\sqrt{10}}-\frac{3}{\sqrt{10}}}\)
\(=\frac{\frac{4}{\sqrt{10}}}{-\frac{2}{\sqrt{10}}}\)
= -2
TH2 :
Cả sin và cos cùng âm
\(\Rightarrow M=\frac{-\frac{1}{\sqrt{10}}+\left(-\frac{3}{\sqrt{10}}\right)}{-\frac{1}{\sqrt{10}}-\left(-\frac{3}{\sqrt{10}}\right)}\)
=\(\frac{-\frac{4}{\sqrt{10}}}{\frac{2}{\sqrt{10}}}\)
= -2
b)
\(B=\frac{sin15+cos15}{cos15}-cot75\)
=\(\frac{sin15}{cos15}+\frac{cos15}{cos15}-cot75\)
=\(tan15+1-cot75\)
=\(cot75+1-cot75\)
= 1
Từ \(\frac{3x+y}{47}=\frac{x+y}{-17}=\frac{-2}{x^2}=\frac{-xz^2-yz^2}{z^2+1}\)(1)
=> \(\frac{x+y}{-17}=\frac{-xz^2-yz^2}{z^2+1}\Rightarrow\frac{x+y}{-17}=\frac{-z^2\left(x+y\right)}{z^2+1}\)
=> (z2 + 1)(x + y) = 17z2(x + y)
=> z2 + 1 = 17z2
=> 16z2 = 1
=> \(z^2=\frac{1}{16}\Rightarrow\orbr{\begin{cases}z=\frac{1}{4}\\z=-\frac{1}{4}\end{cases}}\)
Từ (1) => \(\frac{3x+y}{47}=\frac{x+y}{-17}=\frac{3x+y-x-y}{47+17}=\frac{2x}{64}=\frac{x}{32}\)
Kết hợp với đề bài => \(\frac{x}{32}=\frac{-2}{x^2}\Rightarrow x^3=-64\Rightarrow x=-4\)
\(\frac{3x+y}{47}=\frac{x+y}{-17}\Rightarrow-17\left(3x+y\right)=47\left(x+y\right)\)
=> - 51x - 17y = 47x + 47y
=> -51x - 47x = 17y + 47y
=> -98x = 64y
=> -49x = 32y
=> -49 x (-4) = 32y
=> 196 = 32y
=> y = 6,125
Vậy các cặp (x;y;z) thỏa mãn là (-4 ; 6,125 ; -1/4) ; (-4 ; 6,125 ; 1/4)
a) đkxđ: \(a>0;a\ne1\)
Ta có:
\(P=\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(1-\frac{1}{\sqrt{a}}\right)\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(P=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{\sqrt{a}-1}{\sqrt{a}}.\frac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(P=\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}+\frac{2a+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)
\(P=\frac{2\sqrt{a}\left(\sqrt{a}+1\right)+2a+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)
\(P=\frac{2a+2\sqrt{a}+2a+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)
\(P=\frac{4a+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)
b) \(P=7\)
\(\Leftrightarrow\frac{4a+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\sqrt{a}}=7\)
\(\Leftrightarrow4a+2\sqrt{a}+2=7a+7\sqrt{a}\)
\(\Leftrightarrow3a+5\sqrt{a}-2=0\)
\(\Leftrightarrow\left(3a-\sqrt{a}\right)+\left(6\sqrt{a}-2\right)=0\)
\(\Leftrightarrow\left(3\sqrt{a}-1\right)\sqrt{a}+2\left(3\sqrt{a}-1\right)=0\)
\(\Leftrightarrow\left(3\sqrt{a}-1\right)\left(\sqrt{a}+2\right)=0\)
Mà \(\sqrt{a}+2\ge2\left(\forall a\right)\)
\(\Rightarrow3\sqrt{a}-1=0\Leftrightarrow3\sqrt{a}=1\)
\(\Leftrightarrow\sqrt{a}=\frac{1}{3}\Rightarrow a=\frac{1}{9}\)
Bài 1:
a) \(\left(x+y\right)^2-y^2=x^2+2xy+y^2-y^2=x^2+2xy=x\left(x+2y\right)\)
b) Sửa đề: \(\left(x^2+y^2\right)^2-\left(2xy\right)^2=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)^2\left(x+y\right)^2\)
c) \(x\left(x-3y\right)^2+y\left(y-3x\right)^2=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)\)
\(=x^3-6x^2y+9xy^2+y^3-6xy^2+9x^2y\)
\(=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3\)
Bài 2:
a) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2a\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)\)
\(=2a\left(a^2+3b^2\right)\)
b) \(\left(a+b\right)^3-\left(a-b\right)^3=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2b\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2b\left(b^2+3a^2\right)\)