ta có :

\(\frac{2}{y-2}=\frac{3}{z+2}\Leftrightarrow\frac{2}{y}=\frac{3}{z+5}\Leftrightarrow\frac{4}{y^2}=\frac{9}{\left(z+5\right)^2}\) hay ta có :\(\left(z+5\right)^2=\frac{9}{4}y^2\Rightarrow2y^2-\frac{9}{4}y^2=-25\Leftrightarrow y^2=100\)

TH1.\(y=10\Rightarrow\frac{4}{x+1}=\frac{2}{10-2}=\frac{3}{z+2}\Leftrightarrow\hept{\begin{cases}x=15\\z=10\end{cases}}\)

TH2.\(y=-10\Rightarrow\frac{4}{x+1}=\frac{2}{-10-2}=\frac{3}{z+2}\Leftrightarrow\hept{\begin{cases}x=-25\\z=-20\end{cases}}\)

1.I do => I did

2.do => did

3.goes => is going

4.are => \

1. Tìm lỗi :

1, What did you do last night ?I do my home work => did

2, What home do you get up yesterday morning ? => time

3, Where is your mother now ?She goes shopping => is going

4, Which class you are in  ? I'm in class 6B => are you

(10x + 2⁴) . 7⁴ = 2.7⁷

=> 10x + 16 = 2.7⁷ : 7⁴

=> 10x + 16 = 686

=> 10x = 686 - 16

=> 10x = 670

=> x = 67

cảm ơn Nhạt

Ta có : \(\hept{\begin{cases}\left|\frac{5}{6}x+\frac{7}{8}\right|\ge0\forall x\\\left(\frac{7}{12}:y-\frac{5}{6}\right)^2\ge0\forall y\end{cases}}\Leftrightarrow\hept{\begin{cases}2\left|\frac{5}{6}x+\frac{7}{8}\right|\ge0\\\frac{7}{2}\left(\frac{7}{12}:y-\frac{5}{6}\right)^2\ge0\end{cases}}\)

=> \(2\left|\frac{5}{6}x+\frac{7}{8}\right|+\frac{7}{2}\left(\frac{7}{12}:y-\frac{5}{6}\right)^2\ge0\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\frac{5}{6}x+\frac{7}{8}=0\\\frac{7}{12}:y-\frac{5}{6}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{21}{20}\\y=\frac{7}{10}\end{cases}}\)

Vậy x = -21/20 ; y = 7/10 

a)x/27=-5/6

x=-5/6.27

x=-15

\(a)\)

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

\(\Leftrightarrow f\left(x\right)=-x^3+\left(3x^2-x^2\right)+x-3+2^3\)

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

\(\Leftrightarrow f\left(x\right)=-x^3+2x^2+x+5\)

\(g\left(x\right)=-3x^3-x^2+2x^3+5x-3-4x\)

\(\Leftrightarrow g\left(x\right)=\left(-3x^3+2x^3\right)-x^2+\left(5x-4x\right)-3\)

\(\Leftrightarrow g\left(x\right)=-x^3-x^2+x-3\)

\(b)\)

Theo đề ra: \(h\left(x\right)=f\left(x\right)+g\left(x\right)\)

\(f\left(x\right)=-x^3+2x^2+x+5\)

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

\(\Rightarrow h\left(x\right)=x^2+2x+2\)

tính góc gì thế bạn

ACD^ ah

b. \(\frac{\left(5.20\right)^4}{\left(25.4\right)^5}\)  

= \(\frac{100^4}{100^5}\) 

=\(\frac{1}{100}\)

áp dụng tính chất day tỉ số bằng nhau

thế như nào ý