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

a: x^2-2x+y^2-8y+17=0

=>x^2-2x+1+y^2-8y+16=0

=>(x-1)^2+(y-4)^2=0

=>x=1 và y=4

b: Sửa đề: 4x^2-4xy+y^2+y^2+4y+4=0

=>(2x-y)^2+(y+2)^2=0

=>y=-2 và x=-1

(1)

(x+1)(x-7)+17>0

<=>x^2-6x+9+1>0

<=>(x-3)^2+1>0(dpcm)

..

(7)

-y^2+4y-4-|x+1|≤0

<=>-(y-2)^2-|x+1|≤0

sum 2 so khong duong ko the la so (+)=>dpcm

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

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

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

1.(x+1)(x-7)+17=(x-3)²+1>0

2.-20-(x-5)(x+3)=-34-(x-1)²<0

3.-2(x+3)-(x-2)(x+2)=-(x+1)²-1<0

4.x²+y²+2x+2y+3=(x+1)²+(y+1)²+1>0

5.2x²+2x+y²+2y+5=2(x+1/2)²+(y+1)²+2>0

6.2x²+2y²+2xy+2x+4y+6=(x+y)²+(x+1)²+(y+2)²+1>0

7.-y²+4y-4-/x+1/=-(y-2)²-/x+1/≤0

a) x² + y² + 4x - 10y + 29 = 0

<=> (x² + 4x + 4) + (y² - 10y + 25) = 0

<=> (x+2)² + (y-5)² = 0

Mà: (x+2)² ≥ 0 với mọi x

(y-5)² ≥ 0 với mọi y

=>\(\left\{{}\begin{matrix}\left(x+2\right)^2=0\\\left(y-5\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\y-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=5\end{matrix}\right.\)(T/m)

Vậy x = -2 và y = 5.

b) C = 5x² - 20x + 15

= 5(x² - 4x + 3)

= 5(x² - x - 3x + 3)

= 5[x(x-1) - 3(x-1)]

= 5(x-1)(x-3)

c) x² + y² + 2x - 6y + 10 = 0

<=> (x² + 2x + 1) + (y² - 6y + 9) = 0

<=> (x+1)² + (y-3)² = 0

Mà: (x+1)² ≥ 0 với mọi x

(y-3)² ≥ 0 với mọi y

\(\Rightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\)(T/m)

Vậy x = -1 và y = 3

d) A = 3x² - 12x + 15

= 3(x² - 4x + 5)

= 3(x² - 5x + x - 5)

= 3[x(x-5) + (x-5)]

= 3(x-5)(x+1)

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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