a: Ta có: \(2x^2+y^2-2xy-10x+6y+13=0\)

=>\(y^2-2xy+x^2+6y-6x+x^2-4x+13=0\)

=>\(\left(y-x\right)^2+6\left(y-x\right)+9+x^2-4x+4=0\)

=>\(\left(y-x+3\right)^2+\left(x-2\right)^2=0\)

=>x-2=0 và y-x+3=0

=>x=2 và y=x-3=2-3=-1

b: \(x^2+7y^2-4xy-2x-2y+4=0\)

=>\(x^2-4xy+4y^2-2x+4y+3y^2-6y+4=0\)

=>\(\left(x-2y\right)^2-2\left(x-2y\right)+1+3y^2-6y+3=0\)

=>\(\left(x-2y-1\right)^2+3\left(y-1\right)^2=0\)

=>x-2y-1=0 và y-1=0

=>y=1 và x=2y+1=2*1+1=3

c: \(11x^2+y^2-6xy-14x+2y+9=0\)

=>\(y^2-6xy+9y^2+2y-6x+2x^2-8x+9=0\)

=>\(\left(y-3x\right)^2+2\left(y-3x\right)+1+2x^2-8x+8=0\)

=>\(\left(y-3x+1\right)^2+2\left(x-2\right)^2=0\)

=>x-2=0 và y-3x+1=0

=>x=2 và y=3x-1=3*2-1=5

a) (2x - 5)² - (5 + 2x) = 0

<=> 4x² - 22x + 20 = 0 

\(\Leftrightarrow\left(2x-\dfrac{11}{2}\right)^2=\dfrac{41}{4}\)

\(\Leftrightarrow x=\dfrac{\pm\sqrt{41}+11}{4}\)

b) \(27x^3-54x^2+36x=0\)

\(\Leftrightarrow x\left(3x^2-6x+4\right)=0\)

\(\Leftrightarrow x=0\) (Vì \(3x^2-6x+4=3\left(x-1\right)^2+1>0\forall x\))

c) x³ + 8 - (x + 2).(x - 4) = 0

\(\Leftrightarrow\left(x+2\right).\left(x^2-2x+4\right)-\left(x+2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-3x+8\right)=0\)

\(\Leftrightarrow x=-2\) (Vì \(x^2-3x+8=\left(x-\dfrac{3}{2}\right)^2+\dfrac{23}{4}>0\))

d) \(x^6-1=0\)

\(\Leftrightarrow\left(x^2\right)^3-1=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^4+x^2+1\right)=0\)

\(\Leftrightarrow x^2-1=0\) (Vì \(x^4+x^2+1>0\))

\(\Leftrightarrow x=\pm1\)

\(d,x^6-1=0\\ \Leftrightarrow\left(x^2\right)^3-1^3=0\\ \Leftrightarrow\left(x^2-1\right)\left(x^4+x^2+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x^4+x^2+1=0\left(Vô.lí,vì:x^4\ge0;x^2\ge0,\forall x\in R\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\ c,\left(x^3+8\right)-\left(x+2\right)\left(x-4\right)=0\\ \Leftrightarrow\left(x^3+8\right)-\left(x^2-2x-8\right)=0\\ \Leftrightarrow x^3-x^2+2x+16=0\\ \Leftrightarrow x^3+2x^2-3x^2-6x+8x+16=0\\ \Leftrightarrow x^2\left(x+2\right)-3x\left(x+2\right)+8\left(x+2\right)=0\\ \Leftrightarrow\left(x^2-3x+8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-3x+8=0\left(Vô.lí\right)\\x+2=0\end{matrix}\right.\Leftrightarrow x=-2\)

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2

2:

a: \(3xy^2-3x^3-6xy+3x\)

\(=3x\cdot\left(y^2-2y+1-x^2\right)\)

\(=3x\left\lbrack\left(y-1\right)^2-x^2\right\rbrack\)

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

b: \(3x^2+11x+6\)

\(=3x^2+9x+2x+6\)

=3x(x+3)+2(x+3)

=(x+3)(3x+2)

c: \(-x^3-4xy^2+4x^2y+16x\)

\(=x\left(16+4xy-4y^2-x^2\right)\)

\(=x\cdot\left\lbrack4^2-\left(x^2-4xy+4y^2\right)\right\rbrack=x\cdot\left\lbrack4^2-\left(x-2y\right)^2\right\rbrack\)

=x(4-x+2y)(4+x-2y)

d: \(xz-x^2-yz+2xy-y^2\)

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

=\(z\left(x-y\right)-\left(x-y\right)^2\)

=(x-y)(z-x+y)

e: \(4x^2-y^2-6x+3y\)

=(2x-y)(2x+y)-3(2x-y)

=(2x-y)(2x+y-3)

f: \(x^4-x^3-10x^2+2x+4\)

\(=x^4+2x^3-2x^2-3x^3-6x^2+6x-2x^2-4x+4\)

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

g: \(\left(x^3-x^2+x\right)\left(121-25y^2-10y\right)-\left(x^3-x^2+x\right)-\left(121-25y^2-10y\right)+1\)

\(=\left(x^3-x^2+x\right)\left(121-25y^2-10y-1\right)-\left(121-25y^2-10y-1\right)\)

\(=\left(x^3-x^2+x-1\right)\left\lbrack121-\left(25y^2+10y+1\right)\right\rbrack\)

\(=\left(x-1\right)\left(x^2+1\right)\left\lbrack121-\left(5y+1\right)^2\right\rbrack\)

=(x-1)(x^2+1)(11-5y-1)(11+5y+1)

=(x-1)(x^2+1)(10-5y)(12+5y)

=5(2-y)(x-1)(x^2+1)(5y+12)

Quy đồng mẫu thức các phân thức sau :

a) 

Bài 2:

\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)

\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)

a) x=949/27
    y=755/27
    z=61/9
    các bạn xem giúp mik đúng chx ạ, mik đặt là k

a) x/-3=y/-7=2x/-6=4y/-28=2x+4y/(-6)+(-28)= 68/-34=-2

Vậy x/-3 = -2 => x=(-2)x(-3)=6

       y/-7= -2 => y=(-2)x(-7)=14

      nhớ chọn nhé

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

\(\frac{x}{20}=\frac{y}{9}=\frac{z}{6}=\frac{x-2y+4z}{20-2\cdot9+4\cdot6}=\frac{13}{26}=\frac12\)

=>\(\begin{cases}x=20\cdot\frac12=10\\ y=9\cdot\frac12=\frac92\\ z=6\cdot\frac12=3\end{cases}\)

2: \(\frac{x}{3}=\frac{y}{4}\)

=>\(\frac{x}{15}=\frac{y}{20}\left(1\right)\)

\(\frac{y}{5}=\frac{z}{7}\)

=>\(\frac{y}{20}=\frac{z}{28}\left(2\right)\)

Từ (1),(2) suy ra \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)

mà 2x+3y-z=186

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

\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{2x+3y-z}{2\cdot15+3\cdot20-28}=\frac{186}{62}=3\)

=>\(\begin{cases}x=3\cdot15=45\\ y=3\cdot20=60\\ z=3\cdot28=84\end{cases}\)

3: \(\frac{x}{2}=\frac{2y}{5}=\frac{4z}{7}\)

=>\(\frac{x}{2}=\frac{y}{2,5}=\frac{z}{1,75}\)

mà 3x+5y+7z=123

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

\(\frac{x}{2}=\frac{y}{2,5}=\frac{z}{1,75}=\frac{3x+5y+7z}{3\cdot2+5\cdot2,5+7\cdot1,75}=\frac{123}{30,75}=4\)

=>\(\begin{cases}x=4\cdot2=8\\ y=4\cdot2,5=10\\ z=4\cdot1,75=7\end{cases}\)

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

=>\(\frac{x}{2}=\frac{y}{\frac32}=\frac{z}{\frac43}\)

Đặt \(\frac{x}{2}=\frac{y}{\frac32}=\frac{z}{\frac43}=k\)

=>\(x=2k;y=\frac32k;z=\frac43k\)

xyz=-108

=>\(2k\cdot\frac32k\cdot\frac43k=-108\)

=>\(4k^3=-108\)

=>\(k^3=-27\)

=>k=-3

=>\(\begin{cases}x=2\cdot\left(-3\right)=-6\\ y=\frac32\cdot\left(-3\right)=-\frac92\\ z=\frac43\cdot\left(-3\right)=-4\end{cases}\)