Tìm x:
a) -5.(3x-1)+4.(x-2)=x+7
b) 6x.(x-1)-3.(1-x)=0
K
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S
1
CD
8 tháng 12 2021
a, <=> x2 -2x +1 + 5x -x2 =8
<=> 3x +1 =8
<=> 3x = 7
<=> x= 7/3
b, thiếu đề
c, <=> 2x3 -1 + 2x(4 -x2) = 7
<=> 2x3 + 8x -23 = 8
<=> 8x =8
<=> x=1
TP
3
4 tháng 9 2021
a: Ta có: \(2x^3-18x=0\)
\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b: Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)
\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)
\(\Leftrightarrow-13x=13\)
hay x=-1
c: Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8=3-3x^2\)
\(\Leftrightarrow3x=12\)
hay x=4
4 tháng 9 2021
a) 2x3-18x=0
⇔ 2x(x2-9)=0
⇔ 2x(x-3)(x+3)=0
⇔ \(\left\{{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b)(3x-1)(2x+1)-6x(x+2)=11
⇔ 6x2+x-1-6x2-12x=11
⇔ -11x=12
\(\Leftrightarrow x=-\dfrac{12}{11}\)
c) (x-1)3-(x+2).(x2-2x+4)=3.(1-x2)
⇔ x3-3x2+3x-1-x3-8-3+3x2=0
⇔ 3x=12
⇔ x=4
S
1
25 tháng 12 2021
a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)
I
1
Y
27 tháng 1 2023
\(a,\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=\dfrac{68}{7}-\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=9\)
\(\Leftrightarrow x-\dfrac{1}{3}=3\)
\(\Leftrightarrow x=3+\dfrac{1}{3}\)
\(\Leftrightarrow x=\dfrac{9}{3}+\dfrac{1}{3}\)
\(\Leftrightarrow x=\dfrac{10}{3}\)
\(b,x+30\%x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}.x=-\dfrac{131}{100}\)
\(\Leftrightarrow x.\left(1+\dfrac{3}{10}\right)=-\dfrac{131}{100}\)
\(\Leftrightarrow x.\dfrac{13}{10}=-\dfrac{131}{100}\)
\(\Leftrightarrow x=-\dfrac{131}{100}.\dfrac{10}{13}\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
\(c,-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{2}{10}\)
\(\Leftrightarrow-\dfrac{2}{3}x=-\dfrac{1}{10}\)
\(\Leftrightarrow x=-\dfrac{1}{10}.\left(-\dfrac{3}{2}\right)\)
\(\Leftrightarrow x=\dfrac{3}{20}\)
LJ
2
6 tháng 9 2021
\(a,\sqrt{9x^2}=2x+1\\ \Leftrightarrow\left[{}\begin{matrix}3x=2x+1,\forall x\ge0\\-3x=2x+1,\forall x< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1,\forall x\ge0\left(N\right)\\x=-1,\forall x< 0\left(N\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-1,\forall x+3\ge0\\x+3=1-3x,\forall x+3< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2,\forall x\ge-3\left(N\right)\\x=-\dfrac{1}{2},\forall x< -3\left(L\right)\end{matrix}\right.\Leftrightarrow x=2\)
\(c,\sqrt{x^2-2x+4}=2x-3\left(x\in R\right)\\ \Leftrightarrow x^2-2x+4=\left(2x-3\right)^2\\ \Leftrightarrow x^2-2x+4=4x^2-12x+9\\ \Leftrightarrow3x^2-10x+5=0\\ \Delta=100-4\cdot3\cdot5=40\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10-\sqrt{40}}{6}\\x=\dfrac{10+\sqrt{40}}{6}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5-\sqrt{10}}{3}\\x=\dfrac{5+\sqrt{10}}{3}\end{matrix}\right.\)
KK
6 tháng 9 2021
\(a.\sqrt{9x^2}=2x+1\)
<=> \(\sqrt{9}x=2x+1\)
<=> 3x = 2x + 1
<=> 3x - 2x = 1
<=> x = 1
C
1
30 tháng 8 2021
a: Ta có: \(x\left(2-x\right)+x^2+x=7\)
\(\Leftrightarrow2x-x^2+x^2+x=7\)
\(\Leftrightarrow3x=7\)
hay \(x=\dfrac{7}{3}\)
b: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
19 tháng 1
a: 5-3x=6x+7
=>-3x-6x=7-5
=>-9x=2
=>\(x=-\dfrac{2}{9}\)
b: \(\dfrac{3x-2}{6}-5=3-\dfrac{2\left(x+7\right)}{4}\)
=>\(\dfrac{3x-2}{6}+\dfrac{x+7}{2}=8\)
=>\(\dfrac{3x-2+3\left(x+7\right)}{6}=8\)
=>3x-2+3x+14=48
=>6x+12=48
=>6x=36
=>\(x=\dfrac{36}{6}=6\)
c: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
=>\(\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
=>(x-1)(5x+3-3x+8)=0
=>(x-1)(2x+11)=0
=>\(\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)
d: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
=>\(\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
30 tháng 8 2021
a: Ta có: \(x\left(2-x\right)+\left(x^2+x\right)=7\)
\(\Leftrightarrow2x-x^2+x^2+x=7\)
\(\Leftrightarrow3x=7\)
hay \(x=\dfrac{7}{3}\)
b: Ta có: \(\left(2x+1\right)^2-x\left(4-5x\right)=17\)
\(\Leftrightarrow4x^2+4x+1-4x+5x^2=17\)
\(\Leftrightarrow9x^2=16\)
\(\Leftrightarrow x^2=\dfrac{16}{9}\)
hay \(x\in\left\{\dfrac{4}{3};-\dfrac{4}{3}\right\}\)
DN
2
NA
20 tháng 12 2021
a. 4.(x+41) = 7
x + 41 = 7 : 4 = 1,75
x = 1,75 - 41 = -39,25
b. 4.(x-3) = 72 - 110 = 49 - 1 = 48
x - 3 = 48 : 4 = 12
x = 12 + 3 = 15
a) -5(3x - 1) + 4(x - 2) = x + 7
-15x + 5 + 4x - 8 = x + 7
-11x - 3 = x + 7
-3 = x + 7 + 11x
-3 = 12x + 7
-3 - 7 = 12x
-10 = 12x
x = -10/12 = -5/6
=> x = -5/6
Bài làm
Câu a lm r nên mik lm câu b
b) 6x( x - 1 ) - 3 . ( 1 - x ) = 0
=> 6x( x - 1 ) + 3( x - 1 ) = 0
=> ( 6x + 3 )( x - 1 ) = 0
=> 6x + 3 = 0 hoặc x - 1 = 0
=> x = -1/2 hoặc x = 1
Vậy....