Nếu \(\frac{2x-7}{4x-9}=\frac{7}{9};\frac{3y-6}{4z-8}=\frac{3}{4}\)vàx,y,z khác 0 thì \(\frac{x}{y}+\frac{y}{z}=...\)
Những câu hỏi liên quan
\(\frac{x-1}{2x+3}-\frac{3x+7}{2x-3}=\frac{-4x^2+10}{4x^2-9}\)
\(\frac{x-1}{2x+3}-\frac{3x+7}{2x-3}=\frac{10-4x^2}{4x^2-9}\)
\(\frac{x-1}{2x+3}-\frac{3x+7}{2x-3}=\frac{10-4x^2}{\left(2x-3\right)\left(2x+3\right)}\)
\(\frac{\left(x-1\right)\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}-\frac{\left(3x+7\right)\left(2x+3\right)}{\left(2x+3\right)\left(2x-3\right)}=\frac{10-4x^2}{\left(2x+3\right)\left(2x-3\right)}\)
2x2-3x-2x+3-(6x2+9x+14x+21)=10-4x2
2x2-3x-2x+3-6x2-9x-14x-21=10-4x2
2x2-3x-2x+3-6x2-9x-14x-21-10+4x2=0
2x2-6x2+4x2-3x-2x-9x-14x+3-21-10=0
-28x-28=0
-28x=28
x=28:(-28)
x=-1
\(\frac{x-1}{2x+3}-\frac{3x+7}{2x-3}=\frac{-4x^2+10}{4x^2-9}\)
\(\frac{x-1}{2x+3}-\frac{3x+7}{2x-3}=\frac{-2\left(2x^2-5\right)}{4x^2-9}\)
\(\frac{x-1}{2x+3}-\frac{3x+7}{2x-3}=\frac{-4x^2+10}{\left(2x+3\right)\left(2x-3\right)}\)
\(-4x^2-28x-18=-4x^2+10\)
\(-4x^2-28x-18+4x^2-10=0\)
\(-28x-28=0\)
\(-28x=28\)
\(x=-1\)
Nếu \(\frac{2x-7}{4x-9}=\frac{7}{9}\) , \(\frac{3y-6}{4z-8}=\frac{3}{4}\) và x, y, z khác 0 thì \(\frac{x}{y}+\frac{y}{z}=\)
1, |x-5|=|2x-9|
2, ||2x-11|-x|=8
3, |4x-7|+\(\left|\frac{7-4x}{4x-7}\right|\)=9
4, \(\frac{\left|7x^2-9x+2\right|}{5x+4}\)=2-7x
5, \(\frac{x^2-8x+15}{\left|2x^2-9x-5\right|}\)=3x-9
Với x, y ,z khác 0 và \(\frac{2x-7}{4x-9}=\frac{7}{9},\frac{3y-6}{4z-8}=\frac{3}{4}\Rightarrow\frac{x}{y}+\frac{y}{z}=?\)
\(\frac{x-1}{12}+\frac{2x-12}{14}=\frac{3x-14}{25}+\frac{4x-29}{27}\)\(\frac{3}{x+1}+\frac{4}{x-2}=/frac{5}{x-3}+\frac{6}{x-4}\)\(\frac{2x-1}{2}+\frac{2x+1}{3}=/frac{2x+7}{6}+\frac{2x+9}{7}\)
Xem chi tiết
Nếu\(\frac{2x-7}{4x-9}=\frac{7}{9};\frac{3y-6}{4z-8}=\frac{3}{4}\)và x;y;z khác 0 thì\(\frac{x}{y}+\frac{y}{z}\)=....
giải thích đầy đủ(lưu ý: mình tìm ra x = 0 mà đề cho x\(\ne\)0)
Bài 1 tính
\(\frac{X^2-36}{2X+10}.\frac{3}{6-X}\)
\(\frac{X^2-4}{X^2-9}.\frac{3X+9}{X+2}\)
\(\frac{X^3-8}{5X+20}.\frac{X^2+4X}{X^2+2X+4}\)
\(\frac{4X+12}{\left(X+4\right)^2}:\frac{3X+9}{X+4}\)
\(\frac{5X-10}{X^2+7}:2X+4\)
\(X^2-25:\frac{2X+10}{3X-7}\)
\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}=-\frac{3\left(x+6\right)}{2x+10}=-\frac{3x+18}{2x+10}\)
\(\frac{x^2-4}{x^2-9}\cdot\frac{3x+9}{x+2}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{3\left(x+3\right)}{x+2}=\frac{3\left(x-2\right)}{x-3}\)
\(\frac{x^3-8}{5x+20}\cdot\frac{x^2+4x}{x^2+2x+4}=\frac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}\cdot\frac{x\left(x+4\right)}{x^2+2x+4}=\frac{x\left(x-2\right)}{5}\)
\(\frac{4x+12}{\left(x+4\right)^2}:\frac{3x+9}{x+4}=\frac{4\left(x+3\right)}{\left(x+4\right)^2}\cdot\frac{x+4}{3\left(x+3\right)}=\frac{4}{3\left(x+4\right)}\)
Giải các phương trình sau:
a. \(\frac{4}{2x+3}-\frac{7}{3x-5}=0\)
b. \(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\)
c. \(\frac{2}{2x+1}+\frac{x}{4x^2-1}=\frac{7}{2x-1}\)
d. \(\frac{x^2+5}{25-x^2}=\frac{3}{x+5}+\frac{x}{x-5}\)
\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)
\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)
\(< =>12x-20-14x-21=0\)
\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)
\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)
\(< =>\frac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{4x}{\left(2x-3\right)\left(2x+3\right)}-\frac{2x-3}{\left(2x+3\right)\left(2x-3\right)}=0\)
\(< =>8x+12+4x-2x+3=0\)
\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)
\(\frac{2}{2x+1}+\frac{x}{4x^2-1}=\frac{7}{2x-1}\left(đkxđ:x\ne-\frac{1}{2};\frac{1}{2}\right)\)
\(< =>\frac{2\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\frac{x}{\left(2x+1\right)\left(2x-1\right)}=\frac{7\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\)
\(< =>4x-2+x=14x+7\)
\(< =>14x-5x=-2-7\)
\(< =>9x=-9< =>x=-\frac{9}{9}=-1\left(tm\right)\)
Xem thêm câu trả lời
$\frac{4x+3}{5}$ -$\frac{6x-2}{7}$ =$\frac{5x+4}{3}$ +3
b.
$\frac{x+4}{5}$ -x+4=$\frac{x}{3}$ -$\frac{x-2}{2}$
c.$\frac{5x+2}{6}$ -$\frac{8x-1}{3}$ =$\frac{4x+2}{5}$ -5
d.$\frac{2x+3}{3}$ =$\frac{5-4}{2}$
e. $\frac{5x+3}{12}$ =$\frac{1+2x}{9}$
f.$\frac{7x-1}{6}$ =$\frac{16-x}{5}$
g. $\frac{x-3}{5}$ =6-$\frac{1-2x}{3}$
h. $\frac{3x-2}{6}$ -5=$\frac{3-2(x+7)}{4}$
giúp vs ạ, cần gấp
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
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