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Tìm x . biết :
\(a,\frac{2}{5}:\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}:\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}.\frac{5}{4}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow-x=\frac{1}{2}+\frac{1}{2}\)
\(\Rightarrow-x=1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1\)
a. \(\frac{2}{5}.\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}:\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}.\frac{5}{4}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow-x=\frac{1}{2}+\frac{1}{2}\)
\(\Rightarrow-x=1\)
\(\Rightarrow x=-1\)
Hơi tắt nhá
a) Đặt \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=A\)
\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
mà A\(\le0\)
\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\) phải bằng 0 đê thỏa mãn điều kiện
\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy....
b;c)I hệt câu a nên làm tương tự nhá
d)
Hơi tắt nhá
a) Đặt \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=B\)
B=\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\)
Thay ra ta tính đc :\(z=-\dfrac{11}{20}\)
Vậy....
\(/x-\frac{1}{2}/=\frac{1}{3}\\ =>\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=-\frac{1}{3}\end{cases}}\\ =>\orbr{\begin{cases}x=\frac{1}{3}+\frac{1}{2}\\x=-\frac{1}{3}+\frac{1}{2}\end{cases}}\\ =>\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}\)
\(a,|x-\frac{1}{2}|=\frac{1}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=-\frac{1}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}}\)
\(b,\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\)
\(\frac{28}{27}=x:\frac{3}{7}\)
\(x=\frac{4}{9}\)
a) 3/4 + -1/8 = 5/8
b)-5/12 + -7/24 = -9/8
c) 4/21 - -5/28 = 31/84
d) 1 + -7/28 = 3/4
e) -4/3 - 17/6= -25/6
f) 1/3 - ( 1/2 +1/8 )= -7/24
g)1/21 - ( 1/7 - 1/3 ) = 5/21
h)1/2 - 1/4 + 1/13 + 1/8= 47/104
a) x - 1/10 = 1/15
x=1/15+1/10
x=1/6 Vay x=1/6b) -4/21 - x = -3/7
x=-4/21+3/7 x=5/21 Vay x=5/21c) x + 1/2 = 3/4 - (-1/2)
x+1/2= 5/4
x= 5/4-1/2
x=3/4
Vay x=3/4
d) 4/7 - x = 1/3 - (-2/3)
x= 4/7-1/3-2/3 x= -3/7 Vay x=-3/7
a) 6 - |1/2 - x| = 2/5
|1/2 - x| = 6 - 2/5 = 30/5 - 2/5
|1/2 - x| = 28/5
<=> 1/2 - x = 28/5 hay 1/2 - x = -28/5
<=> x = 1/2 - 28/5 hay x = 1/2 -(-28/5) = 1/2 + 28/5
<=> x = 5/10 - 56/10 hay x = 5/10 + 56/10
<=> x = -51/10 hay x = 61/10
b)|x + 3/5| - 1/2 = 1/2
|x + 3/5| = 1/2 + 1/2
|x + 3/5| = 2/2 = 1
<=> x + 3/5 = 1 hay x + 3/5 = -1
<=> x + 3/5 = 5/5 hay x + 3/5 = -5/5
<=> x = 5/5 - 3/5 hay x = -5/5 - 3/5
<=> x = 2/5 hay x = -8/5
c)4 - |x - 1/5| = -1/2
|x - 1/5| = 4 - (-1/2) = 4 + 1/2
|x - 1/5| = 8/2 + 1/2 = 9/2
<=> x - 1/5 = 9/2 hay x - 1/5 = -9/2
<=> x = 9/2 - 1/5 hay x = -9/2 - 1/5
<=> x = 45/10 - 2/10 hay x = -45/10 - 2/10
<=> x = 43/10 hay x = -47/10
d)|x - 2/5| + 3/4 = 11/4
|x - 2/5| = 11/4 - 3/4
|x - 2/5| = 8/4 = 2
<=> x - 2/5 = 2 hay x - 2/5 = -2
<=> x = 2 + 2/5 hay x = -2 + 2/5
<=> x = 10/5 + 2/5 hay x = -10/5 + 2/5
<=> x = 12/5 hay x = -8/5
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