Tìm x
a/ x/27 = -2/3.6
b/ I x I+ 1/2 =3/4
Tìm x
a. x/15=2/5
b. 3/x-7=27/135
c. 3/4:x + 1/2:1/4=4
a: x=2/5*15=6
b: =>x-7=15
=>x=22
c: =>3/4:x+1/2*4=4
=>3/4:x=4-2=2
=>x=3/8
Tìm X
a) (\(\dfrac{1}{4}\) - X) ( X + \(\dfrac{2}{5}\) ) = 0
b) I 2x + 1 I +\(\dfrac{2}{3}\) = 2
c) (2x - 3 )\(^2\) = 36
d) 7\(^x\) + 2 +2 x 7\(^x\) = 357
a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)
=>\(\left|2x+1\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
c: (2x-3)2=36
=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
d: \(7^{x+2}+2\cdot7^x=357\)
=>\(7^x\cdot49+7^x\cdot2=357\)
=>\(7^x=7\)
=>x=1
a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
\(---\)
b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)
\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)
\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
\(---\)
c) \(\left(2x-3\right)^2=36\)
\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(---\)
d) \(7^{x+2}+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)
\(\Rightarrow7^x\cdot\left(49+2\right)=357\)
\(\Rightarrow7^x\cdot51=357\)
\(\Rightarrow7^x=357:51\)
\(\Rightarrow7^x=7\)
\(\Rightarrow x=1\)
tìm x
a, \(x^2-4x+\dfrac{1}{4}\)
b, \(8x^2-25\)
c, \(x^2+7x=8\)
d, \(x^3-3x=-27+9x\)
e, x(x-3)-7x=-21
f, \(x^3-2x^2+x-2=0\)
g, \(x^2-4x=-4\)
h, \(x^3-x^2+x=-1\)
\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)
\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)
tìm x
a) I x - 3,5 I =7,5
b) I x + 4/5 I - 1/2 = 0
c) 3,6 - I x - 0,4 I = 0
d) I x - 3,5 I + I 4,5 - x I =0
a) \(\Leftrightarrow\left[{}\begin{matrix}x-3,5=7,5\\x-3,5=-7,5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-4\end{matrix}\right.\)
b) \(\Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{2}\\x+\dfrac{4}{5}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{10}\\x=-\dfrac{13}{10}\end{matrix}\right.\)
c) \(\Leftrightarrow\left|x-0,4\right|=3,6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)
d) \(\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\4,5-x=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=4,5\end{matrix}\right.\)(vô lý)
Vậy \(S=\varnothing\)
Tìm x
a) -x/27 -1 = 2/3
b) x - 4 = -14/35 : 20/-21
c) x + 2/3 = -1/12 . -4/5
d) 1/5 - 3/5x = -15/14 :20/-21
e) -3/7x = 3/5 . 28/9
f)1/2x + 3/5x = -2/3
a: =>x/27+1=-2/3
=>x/27=-5/3
=>x=-45
b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)
=>x=221/50
c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)
=>x=1/15-2/3=1/15-10/15=-9/15=-3/5
d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)
=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)
=>x=-37/24
e: =>-3/7x=84/45
=>x=-196/45
f: =>11/10x=-2/3
=>x=-20/33
Bài 2:Tìm x
a) |x-2|-\(\dfrac{3}{5}\)=\(\dfrac{1}{2}\)
b) (x-\(\dfrac{7}{3}\)):\(\dfrac{-1}{3}\)=0,4
c) |x-3|=5
d) (2x+3)\(^2\)=25
e) \(\dfrac{3}{4}\)+\(\dfrac{1}{4}\): x=\(\dfrac{2}{5}\)
f) (x-\(\dfrac{1}{2}\))\(^3\)=\(\dfrac{1}{27}\)
Làm nhanh giúp mik vs ạ(đang rất cần, cảm ơn nhiều ạaa!!!!!)
a)
\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)
c)
\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
d)
\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
e)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)
f)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)
b. Tìm x, y, z biết: (x - 1)/2 = (y + 3)/4 = (z - 5)/6 * v * i +y+z: x + y + z = 27
Tìm x
1) I 3x - 7 I = 2
2) ( 5x - 10 )2 = 100
3) ( -4x + 5 )3 = -27
4) 2004( x - 1 ) ( x + 2 ) = 1
5) 4 ( x + 5 ) + x + 20 = -500
6) 3( x + 15 ) + 2( x + 5 ) = - 175
7) 2( 7 - x ) + 5( 14 - x ) = -136
a/ (ghi lại cái đề)
=>+ 3x-7=2
3x=2+7=9
x=3
+ 3x-7=-2
3x=-2+7=5
x=\(\frac{5}{3}\)
b/ (5x-10)2=100
=> +5x-10=10
5x=10+10=20
x=4
+ 5x-10=-10
5x=-10+10=0
x=0
a/ x=3 và 5/3
b/ (5x-10)2=100
=> +5x-10=10
5x=10+10=20
x=4
+ 5x-10=-10
5x=-10+10=0
x=0
c/(-4x+5)3=-27
-4x+5=-3
-4x=-8
x=2
a)\([x.\dfrac{1}{2}]^{3}=\dfrac{1}{27}\)
b)\([x+\dfrac{1}{2} ]^{2}=\dfrac{4}{5} \)
c) I 3x-4/5 I = 11/5
d) I 2x - 2I = 0
\(a,\left(x.\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x.\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{2}{3}\\ ---\\ b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{5}=\left(\dfrac{2}{\sqrt{5}}\right)^2=\left(-\dfrac{2}{\sqrt{5}}\right)^2 \\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\\x+\dfrac{1}{2}=-\dfrac{2}{\sqrt{5}}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\\x=-\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\end{matrix}\right.\\ Vậy:x=\pm\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\)
\(c,\left|3x-\dfrac{4}{5}\right|=\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}3x-\dfrac{4}{5}=\dfrac{11}{5}\\3x-\dfrac{4}{5}=-\dfrac{11}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=\dfrac{11}{5}+\dfrac{4}{5}=3\\3x=-\dfrac{11}{5}+\dfrac{4}{5}=-\dfrac{7}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{3}=1\\x=-\dfrac{7}{5}:3=-\dfrac{7}{15}\end{matrix}\right.\\ ---\\ d,\left|2x-2\right|=0\\ \Leftrightarrow2x-2=0\\ \Leftrightarrow2x=2\\ \Leftrightarrow x=1\)
a: (x*1/2)^3=1/27
=>x*1/2=1/3
=>x=1/3:1/2=2/3
b: \(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\\x+\dfrac{1}{2}=-\dfrac{2}{\sqrt{5}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{5}}{5}-\dfrac{1}{2}=\dfrac{4\sqrt{5}-5}{10}\\x=\dfrac{-4\sqrt{5}-5}{10}\end{matrix}\right.\)
c: =>3x-4/5=11/5 hoặc 3x-4/5=-11/5
=>3x=3 hoặc 3x=-7/5
=>x=-7/15 hoặc x=1
d: =>2x-2=0
=>2x=2
=>x=1