Bài 1: 

a) Ta có: \(\left(15x^2\cdot y^2\cdot z\right):3xyz\)

\(=\dfrac{15x^2y^2z}{3xyz}\)

\(=5xy\)

b) Ta có: \(3x^2\cdot\left(5x^2-4x+3\right)\)

\(=3x^2\cdot5x^2-3x^2\cdot4x+3x^2\cdot3\)

\(=15x^4-12x^3+9x^2\)

c) Ta có: \(\left(2x^2-3x\right):\left(x-4\right)\)

\(=\dfrac{2x^2-8x+5x-20+20}{x-4}\)

\(=\dfrac{2x\left(x-4\right)+5\left(x-4\right)+20}{x-4}\)

\(=2x+5+\dfrac{20}{x-4}\)

d) Ta có: \(-5xy\cdot\left(3x^2y-5xy+y^2\right)\)

\(=-5xy\cdot3x^2y+5xy\cdot5xy-5xy\cdot y^2\)

\(=-15x^3y^2+25x^2y^2-5xy^3\)

Ta có : \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}\)

Theo tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}=\frac{x^2+3y^2-z^2}{4+27-25}=\frac{22}{6}=\frac{11}{3}\)

\(\Rightarrow x^2=\frac{44}{3}\Rightarrow x=\frac{2\sqrt{11}}{\sqrt{3}}=\frac{2\sqrt{33}}{3}\)

\(\Rightarrow y^2=\frac{99}{3}=33\Rightarrow y=\sqrt{33}\)

\(\Rightarrow z^2=\frac{275}{3}\Rightarrow z=\frac{5\sqrt{33}}{3}\)

g, sửa đề 

 \(5x^2-5xy+7y-7x=5x\left(x-y\right)+7\left(y-x\right)=\left(5x-7\right)\left(x-y\right)\)

h, sửa đề

 \(xy-xz+z-y=x\left(y-z\right)-\left(y-z\right)=\left(x-1\right)\left(y-z\right)\)

i, \(x^3+2x^2-3x-6=x^2\left(x+2\right)-3\left(x+2\right)=\left(x^2-3\right)\left(x+2\right)\)

1. | x + 1| + (y + 2)² = 0
Mà (y + 2)² \(\ge\) 0
Đẳng thức khi . y + 2 \(\ge\) 0
y \(\ge\) - 2
. x + 1 = 0
. x = -1

cảm ơn bn

Bài 1:

a) Ta có: \(2x=5y.\)

=> \(\frac{x}{y}=\frac{5}{2}\)

=> \(\frac{x}{5}=\frac{y}{2}\) và \(x.y=90.\)

Đặt \(\frac{x}{5}=\frac{y}{2}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=2k\end{matrix}\right.\)

Có: \(x.y=90\)

=> \(5k.2k=90\)

=> \(10k^2=90\)

=> \(k^2=90:10\)

=> \(k^2=9\)

=> \(k=\pm3.\)

TH1: \(k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.5=15\\y=3.2=6\end{matrix}\right.\)

TH2: \(k=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).5=-15\\y=\left(-3\right).2=-6\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(15;6\right),\left(-15;-6\right).\)

e) Ta có: \(\frac{x}{y}=\frac{4}{5}.\)

=> \(\frac{x}{4}=\frac{y}{5}\) và \(x.y=20.\)

Đặt \(\frac{x}{4}=\frac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)

Có: \(x.y=20\)

=> \(4k.5k=20\)

=> \(20k^2=20\)

=> \(k^2=20:20\)

=> \(k^2=1\)

=> \(k=\pm1.\)

TH1: \(k=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.4=4\\y=1.5=5\end{matrix}\right.\)

TH2: \(k=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).4=-4\\y=\left(-1\right).5=-5\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(4;5\right),\left(-4;-5\right).\)

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