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a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
\(\Leftrightarrow\frac{13}{36}x=-\frac{8}{45}\)
\(\Rightarrow x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right).\left(-\frac{2}{3}\right)+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow-\frac{4}{9}x+\frac{1}{3}+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{4}{9}x=\frac{77}{60}\)
\(\Rightarrow x=\frac{231}{80}\)
a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
=> \(\frac{4}{9}x-\frac{1}{3}x+\frac{2}{5}-\frac{2}{9}+\frac{1}{4}x=0\)
=> \(\left(\frac{4}{9}x-\frac{1}{3}x+\frac{1}{4}x\right)+\left(\frac{2}{5}-\frac{2}{9}\right)=0\)
=> \(\frac{13}{36}x+\frac{8}{45}=0\)
=> \(\frac{13}{36}x=-\frac{8}{45}\)
=> \(x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}+\frac{1}{5}=\frac{-3}{4}\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}=-\frac{19}{20}\)
=> \(\frac{2}{3}x-\frac{1}{2}=\left(-\frac{19}{20}\right):\left(-\frac{2}{3}\right)=\left(-\frac{19}{20}\right)\cdot\left(-\frac{3}{2}\right)=\frac{57}{40}\)
=> \(\frac{2}{3}x=\frac{57}{40}+\frac{1}{2}=\frac{77}{40}\)
=> \(x=\frac{77}{40}:\frac{2}{3}=\frac{77}{40}\cdot\frac{3}{2}=\frac{231}{80}\)
\(c)\)
\(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=\left(7-\frac{1}{50}+x\right)\)
\(\Rightarrow2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\right)=\left(\frac{350}{50}-\frac{1}{50}+x\right)\)
\(\Rightarrow2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)
\(\Rightarrow2x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)-x=\frac{349}{50}\)
\(\Rightarrow x-\left(1-\frac{1}{50}\right)=\frac{349}{50}\)
\(\Rightarrow x-\frac{49}{50}=\frac{349}{50}\)
\(\Rightarrow x=\frac{349}{50}+\frac{49}{50}\)
\(\Rightarrow x=\frac{199}{25}\)
Vậy \(x=\frac{199}{25}\)
~ Ủng hộ nhé
\(a)2.x-3=x+\frac{1}{2}\)
\(\Rightarrow2x-3-x=\frac{1}{2}\)
\(\Rightarrow x-3=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}+3\)
\(\Rightarrow x=\frac{1}{2}+\frac{6}{2}\)
\(\Rightarrow x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}\)
\(b)4.x-\left(2.x+1\right)=3-\frac{1}{3}+x\)
\(\Rightarrow4.x-2.x-1=\frac{9}{3}-\frac{1}{3}+x\)
\(\Rightarrow2.x-1=\frac{8}{3}+x\)
\(\Rightarrow2x-1-x=\frac{8}{3}\)
\(\Rightarrow x-1=\frac{8}{3}\)
\(\Rightarrow x=\frac{8}{3}+1\)
\(\Rightarrow x=\frac{8}{3}+\frac{3}{3}\)
\(\Rightarrow x=\frac{11}{3}\)
Vậy \(x=\frac{11}{3}\)
~ Ủng hộ nhé
a)\(\frac{1}{2}x+2\frac{1}{2}=3\frac{1}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x=-\frac{3}{4}-\frac{5}{2}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:\left(-3\right)\)
\(\Leftrightarrow x=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\Leftrightarrow x=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\Leftrightarrow x=-\frac{6}{11}\)
biết giải bài 2
x/12=y/14=x.y/12.24=98/288=49/144
=> x/12=49/144=> 49/12
=> y/14=49/144=> 343/72
mới lớp 2 thôi
a) Đặt \(\frac{x}{-2}=\frac{y}{-3}=k\Rightarrow\hept{\begin{cases}x=-2k\\y=-3k\end{cases}}\)
Khi đó 4x - 3y = 9
<=> -8k + 9k = 9
=> k = 9
=> x = -18 ; y = -27
b) Ta có : \(2x=3y\Rightarrow\frac{2x}{6}=\frac{3y}{6}\Rightarrow\frac{x}{2}=\frac{y}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{3}=\frac{x+y}{2+3}=\frac{10}{5}=2\)
=> x = 4 ; y = 6
c) Đặt \(\frac{x}{3}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=3k\\y=4k\end{cases}}\)
Khi đó (3k)2 + (4k)2 = 100
<=> 9k2 + 16k2 = 100
=> 25k2 = 100
=> k2 = 4
=> k = \(\pm\)2
Khi k = 2 => x = 6 ; y = 8
Khi k = -2 => x = -6 ; y = -8
Vậy các cặp (x;y) thỏa mãn cần tìm là (6;8);(-6;-8)
d) Đặt \(\frac{x}{3}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=3k\\y=4k\end{cases}}\)
Khi đó x3 + y3 = 91
<=> (3k)3 + (4k)3 = 91
=> 27k3 + 64k3 = 91
=> 91k3 = 91
=> k3 = 1
=> k = 1
=> x = 3 ; y = 4
e) Đặt \(\frac{x}{5}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=5k\\y=4k\end{cases}}\)
Khi đó x2y = 100
<=> (5k)2.4k = 100
=> 25k2.4k = 100
=> 100k3 = 100
=> k = 1
=> x = 5 ; y = 4
cho thêm điều kiện x,y thuộc Z nữa nhá
\(\frac{3}{x}+\frac{1}{3}=\frac{y}{3}\)
\(\frac{3}{x}=\frac{y-1}{3}\)
\(\Rightarrow x.\left(y-1\right)=9\)
Lập bảng ta có :
| x | 1 | 9 | -1 | -9 | 3 | -3 |
| y-1 | 9 | 1 | -9 | -1 | 3 | -3 |
| y | 10 | 2 | -8 | 0 | 4 | -2 |
Vậy ( x ; y ) = { ( 1 ; 10 ) ; ( 9 ; 2 ) ; ( -1 ; -8 ) ; ( -9 ; 0 ) ; ( 3 ; 4 ) ; ( -3 ; -2 ) }
mấy bài còn lại làm tương tự
a)\(\left|\frac{1}{3}x+\frac{5}{4}\right|-\frac{1}{8}=0\)
\(\Leftrightarrow\left|\frac{1}{3}x+\frac{5}{4}\right|=\frac{1}{8}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}x+\frac{5}{4}=\frac{1}{8}\\\frac{1}{3}x+\frac{5}{4}=-\frac{1}{8}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{27}{8}\\x=-\frac{33}{8}\end{cases}}\)
Vậy x=-27/8 và x=-33/8
b) \(\frac{x-2}{32}=\frac{2}{x-2}\)
\(\Leftrightarrow\left(x-2\right)^2=64\)
\(\Leftrightarrow\left(x-2\right)^2=8^2\)
\(\Leftrightarrow\hept{\begin{cases}x-2=8\\x-2=-8\end{cases}}\Leftrightarrow\hept{\begin{cases}x=10\\x=-6\end{cases}}\)
vậy x=10 hoặc x=-6
a, \(\frac{x-2}{x+1}=\frac{x-3}{x+2}ĐK:x\ne-1;-2\)
\(\Leftrightarrow x^2-4=\left(x-3\right)\left(x+1\right)\Leftrightarrow x^2-4=x^2+x-3x-3\)
\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)
b, \(\frac{2x+1}{x-3}=\frac{2x-3}{x+1}ĐK:x\ne3;-1\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=\left(2x-3\right)\left(x-3\right)\)
\(\Leftrightarrow2x^2+2x+x+1=2x^2-6x-3x+9\)
\(\Leftrightarrow2x^2+3x+1-2x^2+9x-9=0\)
\(\Leftrightarrow12x-8=0\Leftrightarrow x=\frac{2}{3}\)