\(\frac{x}{1}=\frac{4x}{4};\frac{y}{2}=\frac{3y}{6};\frac{z}{3}=\frac{2z}{6}\)

mà \(\frac{x}{1}=\frac{y}{3}=\frac{z}{2}\) nên \(\frac{4x}{4}=\frac{3y}{6}=\frac{2z}{6}\)

áp dụng t/c dãy các tỉ số bằng nhau ta có 

 \(\frac{4x}{4}=\frac{3y}{6}=\frac{2z}{6}=\frac{4x-3y+2z}{4-6+6}=\frac{36}{4}=9\)

nếu \(\frac{x}{1}=9=>x=9\)

bài 4 : Ta có : \(\frac{1+2y}{18}=\frac{1+4y}{24}\left(1\right)\)
\(\Rightarrow24+48y=18+72y \)
\(\Rightarrow y=\frac{1}{4}\)
\(\frac{1+4y}{24}=\frac{1+6y}{6x}\left(2\right)\)
Thay y = \(\frac{1}{4}\) vào (2) ta được x = 5 (thõa mãn )
 

giup di ma cac cau huhu

Ta có VT=\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)

=\(\frac{2\left(x+y+z\right)}{x+y+z}\)=2 (1)   => x+y+z=\(\frac{1}{2}\) <=>  \(y+z=\frac{1}{2}-x\)

                                                                          <=>  \(x+z=\frac{1}{2}-y\)

                                                                           <=> \(x+y=\frac{1}{2}-z\)

Thay \(y+z=\frac{1}{2}-x\)vào (1) ta có:.................................................................

Lúc nào tớ rảnh thì gửi thêm!!!!!!!!!!!!!!!!!!!!!!!!!

a) Áp dụng tính chất dãy tỉ số bằng nhau ta có : \(\frac{x}{20}=\frac{y}{9}=\frac{z}{6}=\frac{x-2y+4z}{20-2.9+4.6}=\frac{13}{26}=\frac{1}{2}\)

* \(\frac{x}{20}=\frac{1}{2}\Rightarrow x=\frac{1}{2}.20=10\)

*\(\frac{y}{9}=\frac{1}{2}\Rightarrow y=\frac{1}{2}.9=\frac{9}{2}\)

*\(\frac{z}{6}=\frac{1}{2}\Rightarrow z=\frac{1}{2}.6=3\)

b)c) đề bn viết ko rõ

a) Đặt \(\frac{x}{5}=\frac{y}{7}=k\)

\(\Rightarrow\hept{\begin{cases}x=5k\\y=7k\end{cases}}\)

\(\Rightarrow xy=5k.7k\)

\(\Rightarrow140=35k^2\)

\(\Rightarrow k^2=4\)

\(\Rightarrow\orbr{\begin{cases}k=2\\k=-2\end{cases}}\)

Với k = 2 ta có :

+) \(\frac{x}{5}=2\Rightarrow x=10\)

+) \(\frac{y}{7}=2\Rightarrow y=14\)

Với k = -2 ta có :

+) \(\frac{x}{5}=-2\Rightarrow x=-10\)

+) \(\frac{y}{7}=-2\Rightarrow y=-14\)

Vậy  \(\left(x;y\right)=\left\{\left(10;14\right);\left(-10;-14\right)\right\}\)

b) Ta có :

\(x:y:z\)\(=\)\(2:5:7\)\(\Rightarrow\)\(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}\)\(\Rightarrow\)\(\frac{3x}{6}=\frac{2y}{10}=\frac{z}{7}\)

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

\(\frac{3x}{6}=\frac{2y}{10}=\frac{z}{7}=\frac{3x+2y-z}{6+10-7}=\frac{27}{9}=3\)

+) \(\frac{x}{2}=3\Rightarrow x=6\)

+) \(\frac{y}{5}=3\Rightarrow y=15\)

+) \(\frac{z}{7}=3\Rightarrow z=21\)

Vậy x = 6, y = 15 và z = 21

_Chúc bạn học tốt_

a, x.y/5.7=140/35

=140/35=4

x/5=4/7

x/7=5/4

x.7=5.4

x.7=20

x=20;7

x=20/7

b,chịu

tk thì tk ko tk cx đc

a, \(\frac{x}{5}=\frac{y}{7}\left(x.y=140\right)\)

Đặt \(\frac{x}{5}=\frac{y}{7}=k\)

\(\Rightarrow7x=5y\)

\(\Rightarrow x.y=7k.5k=35k^2=140\)

\(\Rightarrow k^2=4\Rightarrow k=\pm2\)

\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x=2.7=14\\y=2.5=10\end{cases}}\\\hept{\begin{cases}x=\left(-2\right).7=-14\\y=\left(-2\right).5=-10\end{cases}}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x=2.7=14\\y=2.5=10\end{cases}}\\\hept{\begin{cases}x=\left(-2\right).7=-14\\y=\left(-2\right).5=-10\end{cases}}\end{cases}}\)

Vậy ....

b, \(x:y:z=2:5:7\left(3x+2y-z=27\right)\)

Đặt \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=k\)

\(\Leftrightarrow x=2k;y=5k=z=7k\)

\(\Leftrightarrow3x+2y-z=6k+10k-7k=27\)

\(\Leftrightarrow x=6;y=15;z=21\)

Vậy ...

a)|3x-2|=|3x+5|

x<-5/3 or x>=2/3

3x-2=3x+5=> loai

-5/3<=x<2/3

3x-2=-3x-5

6x=-3;x=-1/2(n)

BÀi 1 : 

\(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=\frac{4x}{4}=\frac{3y}{6}=\frac{2z}{6}\)

Áp dụng dãy tỉ số bằng nhau  ta cso : 

        \(\frac{4x}{4}=\frac{3y}{6}=\frac{2z}{6}=\frac{4x+3y-2z}{4-6+6}=\frac{36}{4}=9\)

=> x = 9.1 = 9

=> y = 9.2 = 18

=> z = 9.3 = 27

I là fans onepiece thì điểm danh☺

Bài 1 câu g bạn kia làm sai mình sửa lại nhá

\(3a^2-6ab+3b^2-12c^2\)

\(=3\left(a^2-2ab+b^2\right)-12c^2\)

\(=3\left(a-b\right)^2-12c^2\)

\(=3\left[\left(a-b\right)^2-4c^2\right]\)

\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)

Để mình làm tiếp cho :))

Bài 2 :

Câu a : \(37,5.8,5-7,5.3,4-6,6.7,5+1,5.37,5\)

\(=\left(37,5.8,5+1,5.37,5\right)-\left(7,5.3,4+6,6.7,5\right)\)

\(=37,5\left(8,5+1,5\right)-7,5\left(3,4+6,6\right)\)

\(=37,5.10-7,5.10\)

\(=10.30=300\)

Câu b : \(35^2+40^2-25^2+80.35\)

\(=\left(35^2+80.35+40^2\right)-25^2\)

\(=\left(30+45\right)^2-25^2\)

\(=75^2-25^2\)

\(=\left(75+25\right)\left(75-25\right)\)

\(=100.50=5000\)

Bài 3 :

Câu a : \(x^3-\dfrac{1}{9}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{1}{9}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{1}{3}\end{matrix}\right.\)

Câu b : \(2x-2y-x^2+2xy-y^2=0\)

\(\Leftrightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)

\(\Leftrightarrow\left(x-y\right)\left(2-x+y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\2-x+y=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x+y=2\Rightarrow x=2-y\end{matrix}\right.\)

Câu c :

\(x\left(x-3\right)+x-3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

\(x^2\left(x-3\right)+27-9x=0\)

\(\Leftrightarrow x^2\left(x-3\right)-9\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\pm3\end{matrix}\right.\)

Bài 4 :

Câu a :

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

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

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

\(=x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(x-1\right)\left(x-3\right)\)

Câu b :

\(x^2+x-6\)

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

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

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

Câu c :

\(x^2-5x+6\)

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

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

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

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

Câu d :

\(x^4+4\)

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

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

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

Bài 1:

a) \(2x^2-2xy-5x+5y\)

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

\(=2x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(2x-5\right)\)

b) \(8x^2+4xy-2ax-ay\)

\(=\left(8x^2+4xy\right)-\left(2ax+ay\right)\)

\(=4x\left(2x+y\right)-a\left(2x+y\right)\)

\(=\left(2x+y\right)\left(4x-a\right)\)

c) \(x^3-4x^2+4x\)

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

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

d) \(2xy-x^2-y^2+16\)

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

\(=-\left[\left(x-y\right)^2-4^2\right]\)

\(=-\left[\left(x-y-4\right)\left(x-y+4\right)\right]\)

e) \(x^2-y^2-2yz-z^2\)

\(=-\left[\left(z^2+2yz+y^2\right)-x^2\right]\)

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

\(=-\left[\left(z+y+x\right)\left(z+y-x\right)\right]\)

g) \(3a^2-6ab+3b^2-12c^2\)

\(=\left(3a^2-6ab+3b^2\right)-12c^2\)

\(=\left(\sqrt{3a}+\sqrt{3b}\right)^2-12c^2\)

\(=\left(\sqrt{3a}+\sqrt{3b}+\sqrt{12c}\right)\left(\sqrt{3a}+\sqrt{3b}-\sqrt{12c}\right)\)