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mình biết làm đấy nhưng không biết ghi vào đây như thế nào!
\(\left|0,2x-3,1\right|+\left|0,2x+3,1\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|0,2x-3,1\right|=0\\\left|0,2x+3,1\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0,2x-3,1=0\\0,2x+3,1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0,2x=3,1\\0,2x=-3,1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=15,5\\x=-15,5\end{matrix}\right.\) (vô lí)
Vậy ko tìm dc x thỏa mãn theo yêu cầu
tìm x,y biết
a, \(\left(x-0,2\right)^{10}+\left(y+3,1\right)^{20}=0\)
b, \(\left(x^2-3^2\right)=16\)
a.
Ta có: (x - 0,2)^10 \(\ge\)0 với mọi x
Ta có: (y+ 3,1)^20 \(\ge\)0 với mọi y
\(\Rightarrow\)( x - 0,2 )^10 = 0 và ( y + 3,1 ) ^20 = 0 (vì chúng cộng lại thì bằng 0 và chúng lớn hơn hoặc bằng 0)
\(\Rightarrow\) ( x - 0,2 ) ^ 10 =0
x - 0,2 = 0
x = 0,2
\(\Rightarrow\)( y + 3,1 ) ^ 20 =0
y + 3,1 = 0
y = - 3,1
Vậy x = 0,2 và y = - 3,1
b, (x^2 - 3^2 )= 16
x^ 2 - 9 =16
x^2 = 25
x^2 = (\(\pm\)5)^2
x = \(\pm\)5.
Vậy x = \(\pm\) 5
mình chỉ biết làm 2 câu b and c thôi bạn thông cảm nha
Tìm x,y,z
b,\(\left(x+\frac{1}{2}\right)^2=\frac{81}{64}\)
Có \(\frac{81}{64}=\left(\frac{9}{8}\right)^2hoặc\frac{81}{64}=\left(-\frac{9}{8}\right)^2\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{9}{8}\right)^2hoặc\left(x+\frac{1}{2}\right)^2=\left(-\frac{9}{8}\right)^2\)
+TH1: \(\left(x+\frac{1}{2}\right)^2=\left(\frac{9}{8}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\frac{9}{8}\)
\(x=\frac{9}{8}-\frac{1}{2}\)
\(x=\frac{9-4}{8}\)
\(x=\frac{5}{8}\)
+TH2:\(\left(x+\frac{1}{2}\right)^2=\left(-\frac{9}{8}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=-\frac{9}{8}\)
\(x=-\frac{9}{8}-\frac{1}{2}\)
\(x=\frac{-9-4}{8}\)
\(x=\frac{-13}{8}\)
Vậy x= \(\frac{5}{8}\)hoặc x=\(\frac{-13}{8}\)
c, \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\) và \(x^2-2y^2+z^2\)
Ta có : \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)\(\Leftrightarrow\)\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}\Rightarrow\frac{x^2}{4}=\frac{2y^2}{18}=\frac{z^2}{25}\)
- Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x^2}{4}=\frac{2y^2}{18}=\frac{z^2}{25}=\frac{x^2-2y^2+z^2}{4-18+25}=\frac{44}{11}=4\)
- Do đó :
\(\frac{x^2}{4}=4\Leftrightarrow\frac{x}{2}=4\Rightarrow x=4.2=8\)
\(\frac{2y^2}{18}=4\Leftrightarrow\frac{y^2}{9}=4\Rightarrow\frac{y}{3}=4\Rightarrow y=4.3=12\)
\(\frac{z^2}{25}=4\Leftrightarrow\frac{z}{5}=4\Rightarrow z=4.5=20\)
vậy x = 8 , y= 12 ,z=20
a) \(\sqrt{3-x}\)=5
=>(\(\sqrt{3-x}\))2=52
=>3-x=25
=>x=-22
a) |0,2x - 3,1| = 6,3
\(0,2x-3,1=\pm6,3\)
Th1:
0,2x - 3,1 = 6,3
0,2x = 6,3 + 3,1
0,2x = 9,4
x = 9,4 : 0,2
x = 47
Th2:
0,2x - 3,1 = - 6,3
0,2x = - 6,3 + 3,1
0,2x = - 3,2
x = - 3,2 : 0,2
x = - 16
Vậy x = 47 hoặc x = - 16
b) |12,1x + 12,1 . 0,1| = 12,1
|12,1(x + 0,1)| = 12,1
\(12,1\left(x+0,1\right)=\pm12,1\)
Th1:
12,1(x + 0,1) = 12,1
x + 0,1 = 1
x = 1 - 0,1
x = 0,9
Th2:
12,1(x + 0,1) = - 12,1
x + 0,1 = - 1
x = - 1 - 0,1
x = - 1,1
Vậy x = 0,9 hoặc x = - 1,1
c) |0,2x - 3,1| + |0,2.x + 3,1| = 0
|0,2x - 3,1| + |0,2x + 3,1| \(\ge\) |0,2x - 3,1 + 0,2x + 3,1| = 0,4x
mà |0,2x - 3,1| + |0,2.x + 3,1| = 0
=> x = 0
x + x : 0,2 = 1,35
x * 1 + x * 5 = 1,35
x * ( 1 + 5 ) = 1,35
x * 6 = 1,35
x = 1,35 : 6
x = 0,225
hok tốt nha ^_^
a)
\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
x + x : 0,2 = 1,35
x * 1 + x * 5 = 1,35
x * ( 1 + 5 ) = 1,35
x * 6 = 1,35
x = 1,35 : 6
x = 0,225
hok tốt nha ^_^
a) \(\left|2,5-x\right|-1,3=0\)
th1: \(2,5-x\ge0\Leftrightarrow x\le2,5\)
\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow2,5-x-1,3=0\Leftrightarrow x=1,2\left(tmđk\right)\)
th2: \(2,5-x< 0\Leftrightarrow x>2,5\)
\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow x-2,5-1,3=0\Leftrightarrow x=3,8\left(tmđk\right)\)
vậy \(x=1,2;x=3,8\)
b) \(1,6.\left|x-0,2\right|=0\Leftrightarrow\left|x-0,2\right|=0\Leftrightarrow x-0,2=0\Leftrightarrow x=0,2\) vậy \(x=0,2\)
c) \(\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\)
th1: \(\dfrac{1}{3}-x\ge0\Leftrightarrow x\le\dfrac{1}{3}\)
\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow\dfrac{1}{3}-x-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{-2}{21}\left(tmđk\right)\)
th2: \(\dfrac{1}{3}-x< 0\Leftrightarrow x>\dfrac{1}{3}\)
\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow x-\dfrac{1}{3}-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{16}{21}\left(tmđk\right)\)
vậy \(x=\dfrac{-2}{21};x=\dfrac{16}{21}\)
d) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
th1: \(x+\dfrac{4}{15}\ge0\Leftrightarrow x\ge\dfrac{-4}{15}\)
\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow x+\dfrac{4}{15}-3,75=-2,15\)
\(\Leftrightarrow x=\dfrac{4}{3}\left(tmđk\right)\)
th2: \(x+\dfrac{4}{15}< 0\Leftrightarrow x< \dfrac{-4}{15}\)
\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow-x-\dfrac{4}{15}-3,75=-2,15\)
\(\Leftrightarrow x=\dfrac{-28}{15}\left(tmđk\right)\)
vậy \(x=\dfrac{4}{3};x=\dfrac{-28}{15}\)
e) ta có : \(\left|x-1,5\right|\ge0\forall x\) và \(\left|2,5-x\right|\ge0\forall x\)
\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|=0\Leftrightarrow\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\) 2 giá trị này khác nhau \(\Rightarrow\) phương trình vô nghiệm
a/ \(\left|12,1x+12,1.0,1\right|=12,1\)
\(\Leftrightarrow\left|12,1.\left(x+0,1\right)\right|=12,1\)
\(\Leftrightarrow\left[{}\begin{matrix}12,1.\left(x+0,1\right)=12,1\\12,1.\left(x+0,1\right)=-12,1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+0,1=1\\x+0,1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0,9\\x=-1,1\end{matrix}\right.\)
Vậy ................
b/ \(\left|0,2x-3,1\right|+\left|0,2x+3,1\right|=0\)
Mà \(\left\{{}\begin{matrix}\left|0,2x-3,1\right|\ge0\\\left|0,2x+3,1\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|0,2x-3,1\right|=0\\\left|0,2x+3,1\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0,2x-3,1=0\\0,2x+3,1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0,2x=3,1\\0,2x=-3,1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=15,5\\x=-15,5\end{matrix}\right.\)
Vậy ..