Tim x:
a) 2x-3=1/2
b) /x+1/=0.25
c) 32/2x=2
d) 64/125=(4/5)^x
e) x/6=-9/18
f) -4/x+3=12/-15
g) x+1/2/0.75=3/2/0.25
K
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13 tháng 8 2023
a: =(6x)^2-(3x-2)^2
=(6x-3x+2)(6x+3x-2)
=(9x-2)(3x+2)
d: \(=\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\)
\(=4x\cdot\left[x^2+2x+1+x^2-2x+1\right]\)
=8x(x^2+1)
e: =(4x)^2-2*4x*3y+(3y)^2
=(4x-3y)^2
f: \(=-\left(\dfrac{1}{4}x^4-2\cdot\dfrac{1}{2}x^2\cdot2y^3+4y^6\right)\)
\(=-\left(\dfrac{1}{2}x^2-2y^3\right)^2\)
g: =(4x)^3+1^3
=(4x+1)(16x^2-4x+1)
k: =x^3(27x^3-8)
=x^3(3x-2)(9x^2+6x+4)
l: =(x^3-y^3)(x^3+y^3)
=(x-y)(x+y)(x^2-xy+y^2)(x^2+xy+y^2)
18 tháng 1
b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
18 tháng 8 2023
`#040911`
`a)`
`6 \times x - 5 = 613`
`=> 6 \times x = 613 + 5`
`=> 6 \times x = 618`
`=> x = 618 \div 6`
`=> x = 103`
Vậy, `x = 103`
`b)`
`12 \times x + 3 \times x = 30`
`=> x \times (12 + 3) = 30`
`=> x \times 15 = 30`
`=> x = 30 \div 15`
`=> x = 2`
Vậy, `x = 2`
`c)`
`125 - 25 \times (x - 1) = 100`
`=> 25 \times (x - 1) = 125 - 100`
`=> 25 \times (x - 1) = 25`
`=> x - 1 = 25 \div 25`
`=> x - 1 = 1`
`=> x = 1 + 1`
`=> x = 2`
Vậy, `x = 2`
`d)`
`(x - 2) \times (9x - 4) = 0?`
`=>`
TH1: `x - 2 = 0`
`=> x = 0 + 2`
`=> x = 2`
TH2: `9x - 4 = 0`
`=> 9x = 4`
`=> x = 4/9`
Vậy, `x \in {2; 4/9}.`
HQ
Hà Quang Minh
Giáo viên
18 tháng 8 2023
\(a,6x-5=613\\ \Leftrightarrow6x=618\\ \Leftrightarrow x=103\\ b,12x+3x=30\\ \Leftrightarrow15x=30\\ \Leftrightarrow x=2\\ c,125-25\left(x-1\right)=100\\ \Leftrightarrow25\left(x-1\right)=25\\ \Leftrightarrow x-1=1\\ \Leftrightarrow x=2\\ d,\left(x-2\right)\cdot\left(9x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{9}\end{matrix}\right.\)
2 tháng 7 2021
a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1
L
19 tháng 5 2022
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
CN
3
LT
6 tháng 2 2021
\(\dfrac{3}{2}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{3}{2}x=-\dfrac{43}{35}\)
\(\Leftrightarrow x=-\dfrac{86}{105}\)
Vậy \(x=-\dfrac{86}{105}\)
\(-\dfrac{11}{12}x+0,25=\dfrac{5}{6}\)
\(\Leftrightarrow-\dfrac{11}{12}x+\dfrac{1}{4}=\dfrac{5}{6}\)
\(\Leftrightarrow-\dfrac{11}{12}x=\dfrac{7}{12}\)
\(\Leftrightarrow x=-\dfrac{7}{11}\)
Vậy \(x=-\dfrac{7}{11}\)
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy x = {3; 1}\(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
6 tháng 2 2021
(x - 2)2 = 1
<=>\(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy x = 3; 1
(2x - 1)3 = -8
<=> 2x - 1 = -2
<=> 2x = -1
<=> x = \(\dfrac{-1}{2}\)
Vậy x = \(\dfrac{-1}{2}\)
27 tháng 2 2021
`a,x(x-1)-(x+2)^2=1`
`<=>x^2-x-x^2-4x-4=1`
`<=>-5x=5`
`<=>x=-1`
`b,(x+5)(x-3)-(x-2)^2=-1`
`<=>x^2+2x-15-x^2+4x-4+1=0`
`<=>6x-18=0`
`<=>x-3=0`
`<=>x=3`
`c,x(2x-4)-(x-2)(2x+3)=0`
`<=>2x(x-2)-(x-2)(2x+3)=0`
`<=>(x-2)(2x-2x-3)=0`
`<=>-3(x-2)=0`
`<=>x-2=0`
`<=>x=2`
`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`
`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`
`<=>4x+26=-12`
`<=>4x=-38`
`<=>x=-19/2`
13 tháng 5 2023
2:
a: =>-2x=10
=>x=-5
b: =>(x-3)(2x+5)=0
=>x=3 hoặc x=-5/2
TÌM X:
a) 2x - 3 = \(\frac{1}{2}\)
2x = \(\frac{1}{2}+3\)
2x = \(\frac{7}{2}\)
x = 2 : \(\frac{7}{2}\)
x = 2 . \(\frac{2}{7}\)
x = \(\frac{4}{7}\)
b) /x+1/ = 0.25
/x+1/ = \(\frac{1}{4}\)
\(\orbr{\begin{cases}x+1=\frac{1}{4}\\x+1=-\frac{1}{4}\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{4}-1\\x=-\frac{1}{4}-1\end{cases}}\)
\(\orbr{\begin{cases}x=-\frac{3}{4}\\x=-\frac{5}{4}\end{cases}}\)
c) 32 : 2x = 2
\(2x=32:2\)
\(2x=16\)
\(x=16:2\)
\(x=8\)
~GOOD STUDY~
a) 2x-3=1/2
=> 2x=1/2+3
=> 2x=7/2
=> x=7/2:2
=> x=7/4
b) |x+1|=0.25
=> \(\orbr{\begin{cases}x+1=0,25\\x+1=-0,25\end{cases}}\)=>\(\orbr{\begin{cases}x=0,25-1\\x=-0,25-1\end{cases}}\)=>\(\orbr{\begin{cases}x=-0,75\\x=-1,25\end{cases}}\)