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\(\frac{x}{-7}=\frac{5}{-35}\)
\(\frac{x.5}{-35}=\frac{5}{-35}\)
=> x . 5 = 5
x = 5 : 5
x = 1
Bài 2 :x+1/3=x-3/4 <=>4.(x+1)=3.(x-3) 4x+4=3x-9 4x-3x=-9-4 x=-13
Bài 1:
ta có: \(\frac{17}{x+1}.\frac{x}{6}=\frac{17x}{6x+6}\)
Để 17x/6x+6 thuộc Z
=> 17x chia hết cho 6x + 6
=> 102x chia hết cho 6x + 6
102x + 102 - 102 chia hết cho 6x + 6
17.(6x+6) - 102 chia hết cho 6x+6
mà 17.(6x+6) chia hết cho 6x + 6
=> 102 chia hết cho 6x + 6
=> ...
bn tự lm típ nha!
Bài 2:
ta có: \(\frac{x+1}{3}=\frac{x-3}{4}\)
\(\Rightarrow4x+4=3x-9\)
\(\Rightarrow4x-3x=-9-4\)
\(x=-13\)
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)
\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)
\(\Rightarrow x=165;y=20;z=25\)
a; 3:\(\frac{2x}{5}\)= 1:0.001
3:\(\frac{2x}{5}\)=1000
\(\frac{2x}{5}\)=1000:3
\(\frac{2x}{5}\)=0.003
2x=0.003.5
2x=0.015
x=0.015:2
x=7.5
a) \(\frac{-2}{5}+\frac{5}{6}.x=\frac{-4}{15}\)
\(\frac{5}{6}.x=\frac{-4}{15}-\frac{-2}{5}\)
\(\frac{5}{6}.x=\frac{2}{15}\)
\(x=\frac{2}{15}:\frac{5}{6}\)
\(x=\frac{4}{25}\)
b) \(\left(x-\frac{1}{5}\right)\left(y+\frac{1}{2}\right)\left(z-3\right)=0\)
\(x-\frac{1}{5}=0\)
\(x=0+\frac{1}{5}\)
\(x=\frac{1}{5}\)
\(\frac{-3}{6}=\frac{x}{-2}=\frac{-9}{y}\)
Ta có \(\frac{-3}{6}=\frac{x}{-2}\Leftrightarrow-3.\left(-2\right)=6.x\Leftrightarrow6=6x\Leftrightarrow x=1\left(TM\right)\)
Ta có \(\frac{1}{-2}=\frac{-9}{y}\Leftrightarrow y=-2.\left(-9\right)\Leftrightarrow y=18\left(TM\right)\)
Vậy \(\hept{\begin{cases}x=1\\y=18\end{cases}}\)
a)ta có xy=7*9=7*3*3
vậy x =9;21 , y=7;3
b) xy=-2*5
mà x<0<y
nên x=-2 ,y=5
c)x-y=5 hay x=y+5
\(\frac{y+5+4}{y-5}=\frac{4}{3}\Rightarrow3y+27=4y-20\Rightarrow y=47\Rightarrow x=52\)
a) \(\frac{1}{x}+\frac{y}{6}=\frac{1}{2}\)
\(\frac{1}{x}=\frac{1}{2}-\frac{y}{6}\)
\(\frac{1}{x}=\frac{3}{6}-\frac{y}{6}\)
\(\frac{1}{x}=\frac{3-y}{6}\)
\(\Rightarrow6=x.\left(3-y\right)\)
Lập bảng ta có :
| 3-y | 2 | 3 | -2 | -3 | 1 | 6 | -1 | -6 |
| x | 3 | 2 | -3 | -2 | 6 | 1 | -6 | -1 |
| y | 1 | 0 | 5 | 6 | 2 | -3 | 4 | 9 |
Vậy ...
b) tương tự câu a
c) \(\frac{x-1}{9}+\frac{1}{3}=\frac{1}{y+2}\)
\(\frac{x-1}{9}+\frac{3}{9}=\frac{1}{y+2}\)
\(\frac{x+2}{9}=\frac{1}{y+2}\)
\(\Rightarrow\left(x+2\right).\left(y+2\right)=9\)
| x+2 | 3 | -3 | 1 | 9 | -1 | -9 |
| y+2 | 3 | -3 | 9 | 1 | -9 | -1 |
| x | 1 | -5 | -1 | 7 | -3 | -11 |
| y | 1 | -5 | 7 | -1 | -11 | -3 |
Vậy ...
d) \(\frac{x}{3}-\frac{4}{y}=\frac{1}{5}\)
\(\frac{4}{y}=\frac{x}{3}-\frac{1}{5}\)
\(\frac{4}{y}=\frac{5x}{15}-\frac{3}{15}\)
\(\frac{4}{y}=\frac{5x-3}{15}\)
\(\Rightarrow4.15=y.\left(5x-3\right)\)
\(\Rightarrow60=y.\left(5x-3\right)\)
Lập bảng ta có :
nhiều tự làm
Mình nhầm đó ạ toán lớp 7 nha mn Ụ w Ụ
a. Vì \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\forall x\\\left|y-\frac{3}{4}\right|\ge0\forall y\\\left|z-1\right|\ge0\forall z\end{cases}}\)=> | x +\(\frac{1}{2}\)| + | y -\(\frac{3}{4}\)| + | z - 1 |\(\ge\)0\(\forall\)x ; y ; z
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|y-\frac{3}{4}\right|=0\\\left|z-1\right|=0\end{cases}}\)<=>\(\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{3}{4}\\z=1\end{cases}}\)
Vậy x = - 1/2 ; y = 3/4 ; z = 1
Câu b,c bạn làm tương tự nhé
@Spirited Away cảm ơn ạ
a) Vì \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\forall x\\\left|y-\frac{3}{4}\right|\ge0\forall y\\\left|z-1\right|\ge0\forall z\end{cases}}\)
=> \(\left|x+\frac{1}{2}\right|+\left|y-\frac{3}{4}\right|+\left|z-1\right|\ge0\forall x,y,z\)
Dấu " = " xảy ra khi và chỉ khi => \(\hept{\begin{cases}x+\frac{1}{2}=0\\y-\frac{3}{4}=0\\z-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{3}{4}\\z=1\end{cases}}\)
Vậy GTNN bằng 0 khi x = -1/2,y = 3/4,z = 1
b) \(\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|=0\)
Vì \(\hept{\begin{cases}\left|x-\frac{3}{4}\right|\ge0\forall x\\\left|\frac{2}{5}-y\right|\ge0\forall y\\\left|x-y+z\right|\ge0\forall x,y,z\end{cases}}\)
=> \(\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|\ge0\forall x,y,z\)
Dấu " = " xảy ra khi và chỉ khi => \(\hept{\begin{cases}\left|x-\frac{3}{4}\right|=0\\\left|\frac{2}{5}-y\right|=0\\\left|x-y+z\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\\left|\frac{3}{4}-\frac{2}{5}+z\right|=0\end{cases}}\)
=> \(\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\\left|\frac{7}{20}+z\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\z=-\frac{7}{20}\end{cases}}\)
Vậy GTNN bằng 0 khi x = 3/4,y = 2/5,z = -7/20
c) Vì \(\left|x-\frac{2}{3}\right|\ge0\forall x;\left|x+y+\frac{3}{4}\right|\ge0\forall x,y;\left|y-z-\frac{5}{6}\right|\ge0\forall y,z\)
=> \(\left|x-\frac{2}{3}\right|+\left|x+y+\frac{3}{4}\right|+\left|y-z-\frac{5}{6}\right|\ge0\forall x,y,z\)
Dấu " = " xảy ra khi và chỉ khi \(\hept{\begin{cases}\left|x-\frac{2}{3}\right|=0\\\left|x+y+\frac{3}{4}\right|=0\\\left|y-z-\frac{5}{6}\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{2}{3}\\\left|\frac{2}{3}+y+\frac{3}{4}\right|=0\\\left|y-z-\frac{5}{6}\right|=0\end{cases}}\)
=> \(\hept{\begin{cases}x=\frac{2}{3}\\\left|\frac{17}{12}+y\right|=0\\\left|y-z-\frac{5}{6}\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=-\frac{17}{12}\\\left|-\frac{17}{12}-z-\frac{5}{6}\right|=0\end{cases}}\)
=> \(\hept{\begin{cases}x=\frac{2}{3}\\y=-\frac{17}{12}\\z=-\frac{9}{4}\end{cases}}\)
Vậy GTNN bằng 0 khi x = 2/3,y = -17/12,z = -9/4
b) Ta có: \(\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|\ge0\left(\forall x,y,z\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|x-\frac{3}{4}\right|=0\\\left|\frac{2}{5}-y\right|=0\\\left|x-y+z\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\\frac{3}{4}-\frac{2}{5}+z=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\z=-\frac{7}{20}\end{cases}}\)
cảm ơn mn nhiều ạ <3
c) Vì \(\left|x-\frac{2}{3}\right|+\left|x+y+\frac{3}{4}\right|+\left|y-z-\frac{5}{6}\right|\ge0\left(\forall x,y,z\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|x-\frac{2}{3}\right|=0\\\left|x+y+\frac{3}{4}\right|=0\\\left|y-z-\frac{5}{6}\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\\frac{2}{3}+y+\frac{3}{4}=0\\y-z-\frac{5}{6}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=-\frac{17}{12}\\-\frac{17}{12}-z-\frac{5}{6}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=-\frac{17}{12}\\z=\frac{9}{4}\end{cases}}\)
a) Vì \(\left|x+\frac{1}{2}\right|+\left|y-\frac{3}{4}\right|+\left|z-1\right|\ge0\left(\forall x,y,z\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|y-\frac{3}{4}\right|=0\\\left|z-1\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{3}{4}\\z=1\end{cases}}\)