Bài 7: 

\(\widehat{C}=180^0-70^0-40^0=70^0=\widehat{B}\)

Ta có:

\(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=4\)

\(\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+4=4\)

\(\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2\ge0\forall x\\\left(y-\dfrac{1}{y}\right)\ge0\forall y\end{matrix}\right.\)

Dấu "="⇔ \(\left\{{}\begin{matrix}x=\dfrac{1}{x}\\y=\dfrac{1}{y}\end{matrix}\right.\)

\(\Leftrightarrow x^2=y^2=1\)

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

Thay vào phương trình 1

⇒ \(x=y=1\)

2:

Chiều cao là: 26*2:8=6,5cm

S2-S1=1/2*6,5*11-1/2*6,5*8=9,75cm²

\(A=\frac{3n-7}{n+2}=\frac{3.\left(n+2\right)-13}{n+2}=3-\frac{13}{n+2}\inℤ\)

\(\Rightarrow\frac{13}{n+2}\inℤ\)

Mà \(13\inℤ\Rightarrow n+2\inℤ\Rightarrow n\inℤ\)và \(n+2\inƯ\left(13\right)\)

\(\Rightarrow\left(n+2\right)\in\left\{\pm1;\pm13\right\}\)\(\Rightarrow n\in\left\{-1;-3;11;-15\right\}\)

tự chép đi. thế thì tự đi mà trả lời

𝑦+5.684.3=7.6

\frac{y+5.68}{4.3}=7.64.3y+5.68​=7.6

43(𝑦+5.684310)=43⋅7.6

Tìm \(y\)

\(\left(y+5,68\right):4,3=7,6\)

\(y+5,68=7,6\times4,3\)

\(y+5,68=32,68\)

\(y=32,68-5,68\)

\(y=27\)

( y + 5,68 ) : 4,3 = 7,6

y + 5,68 = 7,6 x 4,3

y + 5,68 = 32,68

y = 32,68 - 5,68

y = 27

\(37.46+37.54-5^3.2^3\\ =37\left(46+54\right)-125.8\\ =37.100-1000\\ =3700-1000\\ =2700\)

\(37\cdot46+37\cdot54-5^3\cdot2^3\)
\(\Rightarrow37\cdot\left(46+54\right)-125\cdot8\)
\(\Rightarrow37\cdot100-1000\)
\(\Rightarrow3700-1000\)
\(\Rightarrow2700\)

=37.46+37.54-125.8

=37(46+54)-1000

=37.100-1000

=2700