I. Tìm x:
a. 3.(x+1) - 5 = 2x + 7
b. x - 2 = 6.(x - 1) + 1
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I
1
Y
27 tháng 1 2023
\(a,\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=\dfrac{68}{7}-\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=9\)
\(\Leftrightarrow x-\dfrac{1}{3}=3\)
\(\Leftrightarrow x=3+\dfrac{1}{3}\)
\(\Leftrightarrow x=\dfrac{9}{3}+\dfrac{1}{3}\)
\(\Leftrightarrow x=\dfrac{10}{3}\)
\(b,x+30\%x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}.x=-\dfrac{131}{100}\)
\(\Leftrightarrow x.\left(1+\dfrac{3}{10}\right)=-\dfrac{131}{100}\)
\(\Leftrightarrow x.\dfrac{13}{10}=-\dfrac{131}{100}\)
\(\Leftrightarrow x=-\dfrac{131}{100}.\dfrac{10}{13}\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
\(c,-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{2}{10}\)
\(\Leftrightarrow-\dfrac{2}{3}x=-\dfrac{1}{10}\)
\(\Leftrightarrow x=-\dfrac{1}{10}.\left(-\dfrac{3}{2}\right)\)
\(\Leftrightarrow x=\dfrac{3}{20}\)
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: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
25 tháng 10 2021
a) \(=2x^2-7x-15-2x^2+6x+x+7=-8\)
b) \(=2x^2+x-x^3-2x^2+x^3-x+3=3\)
30 tháng 9 2021
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
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 \(x\in\left\{\dfrac{4}{3};-\dfrac{4}{3}\right\}\)
NM
2
28 tháng 8 2021
a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)
\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)
\(\Leftrightarrow2x=-8\)
hay x=-4
b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)
\(\Leftrightarrow-10x=-10\)
hay x=1
c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)
\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)
\(\Leftrightarrow-4x=-8\)
hay x=2
3 tháng 10 2021
a: Ta có: \(\left(x-5\right)\left(x+3\right)=x\left(x-3\right)\)
\(\Leftrightarrow x^2-2x-15-x^2+3x=0\)
\(\Leftrightarrow x=15\)
b: Ta có: \(\left(x+2\right)^2=\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow x+2=0\)
hay x=-2
c: Ta có: \(\left(x-6\right)\left(x+6\right)=x^2\)
\(\Leftrightarrow x^2-36=x^2\)(vô lý)
HP
3 tháng 10 2021
a. (x - 5)(x + 3) = x(x - 3)
<=> x2 + 3x - 5x - 15 = x2 - 3x
<=> x2 - x2 + 3x - 5x + 3x - 15 = 0
<=> x = 15
b. (x + 2)2 = (x - 1)(x + 2)
<=> x2 + 4x + 4 = x2 + 2x - x - 2
<=> x2 - x2 + 4x - 2x + x = -2 - 4
<=> 3x = -5
<=> \(x=\dfrac{-5}{3}\)
c. (x - 6)(x + 6) = x2
<=> x2 - 36 - x2 = 0
<=> x2 - x2 = 36
<=> 0 = 36 (vô lí)
Vậy nghiệm của PT là \(S=\varnothing\)
d. (2x - 3)2 = 4x2 - 8
<=> 4x2 - 12x + 9 - 4x2 + 8 = 0
<=> 4x2 - 4x2 - 12x = -8 - 9
<=> -12x = -17
<=> \(x=\dfrac{17}{12}\)
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) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
C
2
21 tháng 8 2021
a: Ta có: \(\left(2x-3\right)^2+6\left(2x-1\right)=7\)
\(\Leftrightarrow\left(2x-3\right)^2+6\left(2x-1\right)-7=0\)
\(\Leftrightarrow4x^2-12x+9+12x-6-7=0\)
\(\Leftrightarrow4x^2=4\)
\(\Leftrightarrow x^2=1\)
hay \(x\in\left\{1;-1\right\}\)
b: Ta có: \(x^2-7x+10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
C
4
21 tháng 8 2021
a) \(\left(2x-3\right)^2+6\left(2x-1\right)=7\\ \Rightarrow4x^2-12x+9+12x-6-7=0\\ \Rightarrow4x^2-4=0\\ \Rightarrow x^2-1=0\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
b) \(x^2-7x+10=0\\ \Rightarrow\left(x^2-2x\right)-\left(5x-10\right)=0\\ \Rightarrow\left(x-2\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
c) \(-6x^2+13x-5=0\\ \Rightarrow-\left(6x^2-13x+5\right)=0\\ \Rightarrow-\left[\left(6x^2-10x\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left[2x\left(3x-5\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left(2x-1\right)\left(3x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\left(2x-1\right)=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\)
a, \(3\left(x+1\right)-5=2x+7\)
\(\Leftrightarrow3x+3-5=2x+7\Leftrightarrow x=9\)
b, \(x-2=6\left(x-1\right)+1\)
\(\Leftrightarrow x-2=6x-6+1\Leftrightarrow-5x=-3\Leftrightarrow x=\frac{3}{5}\)
a) 3( x + 1 ) - 5 = 2x + 7
<=> 3x + 3 - 5 = 2x + 7
<=> 3x - 2x = 7 - 3 + 5
<=> x = 9
b) x - 2 = 6( x - 1 ) + 1
<=> x - 2 = 6x - 6 + 1
<=> x - 6x = -6 + 1 + 2
<=> -5x = -3
<=> x = 3/5