\(5x=3y\Rightarrow x=\dfrac{3y}{5}\)

Thay \(x=\dfrac{3y}{5}\) vào biểu thức \(x^2-y^2=-4\) ta có:

\(\left(\dfrac{3y}{5}\right)^2-y^2=-4\)

\(\dfrac{9y^2}{25}-y^2=-4\)

\(-\dfrac{16}{25}y^2=-4\)

\(y^2=-\dfrac{4}{\dfrac{-16}{25}}\)

\(y^2=\dfrac{25}{4}\)

\(\Rightarrow y=-\dfrac{5}{2};y=\dfrac{5}{2}\)

*) \(y=-\dfrac{5}{2}\Rightarrow x=\dfrac{3.\left(-\dfrac{5}{2}\right)}{5}=-\dfrac{3}{2}\)

*) \(y=\dfrac{5}{2}\Rightarrow x=\dfrac{3.\dfrac{5}{2}}{5}=\dfrac{3}{2}\)

Vậy ta được các cặp giá trị \(\left(x;y\right)\) thỏa mãn:

\(\left(-\dfrac{3}{2};-\dfrac{5}{2}\right);\left(\dfrac{3}{2};\dfrac{5}{2}\right)\)

ai biết gì đâu

Lời giải:
Áp dụng tính chất tổng 3 góc trong một tam giác bằng $180^0$

a.

$x=180^0-80^0-45^0=55^0$

b.

$y=180^0-30^0-90^0=60^0$

c.

$z=180^0-30^0-25^0=125^0$

Bài 3:

a: \(\frac{31}{15}>1;\frac{15}{31}<1\)

Do đó: \(\frac{31}{15}>\frac{15}{31}\)

=>\(\left(\frac{31}{15}\right)^{11}>\left(\frac{15}{31}\right)^{11}\)

b: \(\frac89<1\)

=>\(\left(\frac89\right)^{23}>\left(\frac89\right)^{25}\)

=>\(-\left(\frac89\right)^{23}<-\left(\frac89\right)^{25}\)

=>\(\left(-\frac89\right)^{23}<\left(-\frac89\right)^{25}\)

c: \(27^{40}=\left(27^2\right)^{20}=729^{20}\)

\(64^{60}=\left(64^3\right)^{20}=262144^{20}\)

mà 729<262144

nên \(27^{40}<64^{60}\)

Bài 2:

a: \(A=\frac{1}{10}-\frac{1}{10\cdot9}-\frac{1}{9\cdot8}-\cdots-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)

\(=\frac{1}{10}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{9\cdot10}\right)\)

\(=\frac{1}{10}-\left(1-\frac12+\frac12-\frac13+\cdots+\frac19-\frac{1}{10}\right)\)

\(=\frac{1}{10}-\left(1-\frac{1}{10}\right)=\frac{1}{10}-\frac{9}{10}=-\frac{8}{10}=-\frac45\)

b: \(B=\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)

=>\(3B=1+\frac13+\cdots+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)

=>\(3B-B=1+\frac13+\cdots+\frac{1}{3^{98}}+\frac{1}{3^{99}}-\frac13-\frac{1}{3^2}-\cdots-\frac{1}{3^{100}}\)

=>\(2B=1-\frac{1}{3^{100}}=\frac{3^{100}-1}{3^{100}}\)

=>\(B=\frac{3^{100}-1}{2\cdot3^{100}}\)

Bài 2:

DE//BC

=>\(\hat{ADE}=\hat{ABC}\) (hai góc đồng vị)

mà \(\hat{ABC}=80^0\)

nên \(\hat{ADE}=80^0\)

Ta có: DE//BC

=>\(\hat{AED}=\hat{ACB}\) (hai góc đồng vị)

=>\(\hat{AED}=60^0\)

cho bức hình bé tí thế

a: \(\left(-\frac54x+3,25\right)\left\lbrack\frac35-\left(-\frac52x\right)\right\rbrack=0\)

=>\(\left(\frac54x-\frac{13}{4}\right)\left(\frac52x+\frac35\right)=0\)

=>\(\left[\begin{array}{l}\frac54x-\frac{13}{4}=0\\ \frac52x+\frac35=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac54x=\frac{13}{4}\\ \frac52x=-\frac35\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{13}{4}:\frac54=\frac{13}{5}\\ x=-\frac35:\frac52=-\frac{6}{25}\end{array}\right.\)

b: \(\left(-\frac72x+1,75\right)\left\lbrack\frac45-\left(-\frac53x\right)\right\rbrack=0\)

=>\(\left[\begin{array}{l}-\frac72x+1,75=0\\ \frac45-\left(-\frac53x\right)=0\end{array}\right.\Longrightarrow\left[\begin{array}{l}-\frac72x=-1,75=-\frac74\\ \frac53x=-\frac45\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{-7}{4}:\frac{-7}{2}=\frac24=\frac12\\ x=-\frac45:\frac53=-\frac45\cdot\frac35=-\frac{12}{25}\end{array}\right.\)

c: \(\left(x^2-4\right)\left(x+\frac27\right)=0\)

=>\(\left[\begin{array}{l}x^2-4=0\\ x+\frac27=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=4\\ x=-\frac27\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-2\\ x=-\frac27\end{array}\right.\)

d: \(\left(25-x^2\right)\left(5x-\frac59\right)=0\)

=>\(\left[\begin{array}{l}25-x^2=0\\ 5x-\frac59=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=25\\ 5x=\frac59\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-5\\ x=\frac19\end{array}\right.\)