a: Ta có: \(x^2-8x+20\)

\(=x^2-8x+16+4\)

\(=\left(x-4\right)^2+4>0\forall x\)

b: Ta có: \(-x^2+6x-19\)

\(=-\left(x^2-6x+19\right)\)

\(=-\left(x^2-6x+9+10\right)\)

\(=-\left(x-3\right)^2-10< 0\forall x\)

<=> 32( \(x^2-\frac{3}{16}-\frac{13}{16}\)) = 0

<=> 32\(\left(x^2-x+\frac{13}{16}x-\frac{13}{16}\right)\)= 0

<=> 32[ x(x-1)+ 13/16( x- 1)] = 0

<=> 32( x-1) (x+13/16)= 0

<=> x-1=0  hoặc  x+13/16= 0

<=> x= 1 hoặc  x= -13/16

32x^2 - 6x - 26 = 0

<=> (x- 1) (x+\(\frac{13}{16}\)) = 0

<=> x= 1 hoặc  x= \(\frac{-13}{16}\)

bài nay bn p dùng máy tính, ko thì ko lm đc đâu

sao lại ra 13/13 z ạ

1.

PT $\Leftrightarrow (x^2+2xy+y^2)-(y^2+6y+9)=0$

$\Leftrightarrow (x+y)^2-(y+3)^2=0$

$\Leftrightarrow (x+y-y-3)(x+y+y+3)=0$

$\Leftrightarrow (x-3)(x+2y+3)=0$

$\Rightarrow x-3=0$ hoặc $x+2y+3=0$

Nếu $x-3=0\Leftrightarrow x=3$. Vậy $(x,y)=(3,a)$ với $a$ nguyên bất kỳ.

Nếu $x+2y+3=0\Leftrightarrow x=-2y-3$ lẻ. Vậy $(x,y)=(-2a-3,a)$ với $a$ nguyên bất kỳ.

2. 

PT $\Leftrightarrow x^2=(y^2+2y+1)+12$

$\Leftrightarrow x^2=(y+1)^2+12\Leftrightarrow x^2-(y+1)^2=12$

$\Leftrightarrow (x-y-1)(x+y+1)=12$
Vì $x-y-1, x+y+1$ là số nguyên và cùng tính chẵn lẻ nên xảy ra các TH sau:

TH1: $x-y-1=2; x+y+1=6\Rightarrow x=4; y=1$

TH2: $x-y-1=6; x+y+1=2\Rightarrow x=4; y=-3$

TH3: $x-y-1=-2; x+y+1=-6\Rightarrow x=-4; y=-3$

TH4: $x-y-1=-6; x+y+1=-2\Rightarrow x=-4; y=1$

ta co : \(\frac{x}{2}=\frac{y}{3}:\frac{y}{5}=\frac{z}{4}\)

=> \(\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{12}\)

=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\) va : x - y + z = -49

AD tinh chat day ti so = nhau ta co :

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{-49}{7}=-7\)

\(\frac{x}{10}=-7=>x=-7.10=-70\)

\(\frac{y}{15}=-7=>y=15.-7=-105\)

\(\frac{z}{12}=-7=>z=12.-7=-84\)

vay : x = -70 : y = -105 ; z = -84

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\) (1)

\(\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{15}=\frac{z}{12}\) (2)

Từ (1) và (2)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{-49}{7}=-7\)

\(\frac{x}{10}=-7\Rightarrow x=-7\times10=-70\)

\(\frac{y}{15}=-7\Rightarrow y=-7\times15=-105\)

\(\frac{z}{12}=-7\Rightarrow z=-7\times12=-84\)

Ta có : a mũ chẵn \(\ge\)0.

=>\(2\times y-8=0\)

=> 2 x y = 8

=> y = 4

Ta có : 2x-y = 0.

=> 2x=y=8

=>x= 4

x = 2 nhé.

mk biet roi 

Xét \(x^3-x^2+x-5=0\)

\(\Leftrightarrow\left(x-\frac{1}{3}\right)^3+\frac{2}{3}\left(x-\frac{1}{3}\right)=\frac{128}{27}\)

Xét \(y^3-2y^2+2y+4=0\)

\(\Leftrightarrow\left(y-\frac{2}{3}\right)^3+\frac{2}{3}\left(y-\frac{2}{3}\right)=-\frac{128}{27}\)

Cộng theo vế 2 dòng có dấu <=> ta có:

\(\left(x-\frac{1}{3}\right)^3+\left(y-\frac{2}{3}\right)^3+\frac{2}{3}\left(x-\frac{1}{3}+y-\frac{2}{3}\right)=0\)

\(\Leftrightarrow\left(x-\frac{1}{3}+y-\frac{2}{3}\right)\left(\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\left(y-\frac{2}{3}\right)+\left(y-\frac{2}{3}\right)^2\right)+\frac{2}{3}\left(x+y-1\right)=0\)

\(\Leftrightarrow\left(x+y-1\right)\left(\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\left(y-\frac{2}{3}\right)+\left(y-\frac{2}{3}\right)^2\right)+\frac{2}{3}\left(x+y-1\right)=0\)

\(\Leftrightarrow\left(x+y-1\right)\left(\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\left(y-\frac{2}{3}\right)+\left(y-\frac{2}{3}\right)^2+\frac{2}{3}\right)=0\)

Dễ thấy: \(\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\left(y-\frac{2}{3}\right)+\left(y-\frac{2}{3}\right)^2+\frac{2}{3}>0\)

\(\Rightarrow x+y-1=0\Rightarrow x+y=1\)

Done !!!

sai hdt roi ban oi

\(5\left(x+2\right)-x^2-2x=0\)

\(\Rightarrow5\left(x+2\right)-\left(x^2+2x\right)=0\)

\(\Rightarrow5\left(x+2\right)-x\left(x+2\right)=0\)

\(\Rightarrow\left(5-x\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5-x=0\\x+2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\)

khó phết