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12 tháng 6 2019
a) \(2x\left(x-3\right)-x\left(2x+1\right)-3\left(x+5\right)=11\)
\(\Rightarrow2x^2-6x-2x^2-x-3x-15=11\)
\(\Rightarrow-10x=26\Rightarrow x=-2,6\)
Vậy ...........
b) \(x\left(x-1\right)-\left(x^2+3x-5\right)-2\left(x+3\right)=17\)
\(\Rightarrow x^2-x-x^2-3x+5-2x-6=17\)
\(\Rightarrow-6x=18\Rightarrow x=-3\)
c) \(5x\left(x-7\right)-\left(5x+1\right)x-\left(x+3\right)2=13\)
\(\Rightarrow5x^2-35x-5x^2-x-2x-6=13\)
\(\Rightarrow-38x=19\Rightarrow x=-\frac{1}{2}\)
d) \(\left(2x^2-3x+5\right)-2x\left(x-3\right)+\left(x-1\right)\left(-2\right)=10\)
\(\Rightarrow2x^2-3x+5-2x^2+6x-2x+2=10\)
\(\Rightarrow x=3\)
CT
0
3 tháng 8 2018
a) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow\left(24x^2+16x-9x-6\right)-\left(4x^2+16x+7x+28\right)=10x^2-2x+5x-1\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-16x-7x-28=10x^2-2x+5x-1\)
\(\Leftrightarrow24x^2+16x-9x-4x^2-16x-7x-10x^2+2x-5x=6+28-1\)
\(\Leftrightarrow10x^2-19x=33\)
\(\Leftrightarrow10x^2-19x+33=0\)
Phương trình vô nghiệm!!!!!!!!
b) \(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)=3\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow4\left(x^2+5x-x-5\right)-\left(x^2+5x+2x+10\right)=3\left(x^2+2x-x-2\right)\)
\(\Leftrightarrow4x^2+20x-4x-20-x^2-5x-2x-10=3x^2+6x-3x-6\)
\(\Leftrightarrow4x^2+20x-4x-x^2-5x-2x-3x^2-6x+3x=20+10-6\)
\(\Leftrightarrow6x=24\)
\(\Leftrightarrow x=4\)
Vậy \(S=\left\{4\right\}\)
24 tháng 9 2020
a) \(x^3=x^5\)
=> \(x^3-x^5=0\)
=> \(x^3\left(1-x^2\right)=0\)
=> \(\orbr{\begin{cases}x^3=0\\1-x^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(4x\left(x+1\right)=x+1\)
=> \(4x^2+4x-x-1=0\)
=> \(4x\left(x+1\right)-1\left(x+1\right)=0\)
=> \(\left(x+1\right)\left(4x-1\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{4}\end{cases}}\)
c) \(x\left(x-1\right)-2\left(1-x\right)=0\)
=> \(x\left(x-1\right)-\left[-2\left(x+1\right)\right]=0\)
=> \(x\left(x-1\right)+2\left(x-1\right)=0\)
=> \(\left(x-1\right)\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
d) Kết quả ?
e) \(\left(x-3\right)^2+3-x=0\)
=> \(x^2-6x+9+3-x=0\)
=> \(x^2-7x+12=0\)
=> \(x^2-3x-4x+12=0\)
=> \(x\left(x-3\right)-4\left(x-3\right)=0\)
=> (x - 4)(x - 3) = 0
=> \(\orbr{\begin{cases}x=4\\x=3\end{cases}}\)
f) Tương tự
KN
13 tháng 8 2020
a. \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)
\(\Rightarrow10x^2-35x+16x-10x^2=5\)
\(\Rightarrow-19x=5\)
\(\Rightarrow x=-\frac{5}{19}\)
b. \(x\left(x-\frac{1}{3}\right)-\frac{1}{2}x\left(2x-3\right)=\frac{1}{4}\)
\(\Rightarrow x^2-\frac{1}{3}x-x^2+\frac{3}{2}x=\frac{1}{4}\)
\(\Rightarrow\frac{7}{6}x=\frac{1}{4}\)
\(\Rightarrow x=\frac{3}{14}\)
c. \(5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)
\(\Rightarrow5x^2-15x+5+x-5x^2=x-2\)
\(\Rightarrow-14x+5=x-2\)
\(\Rightarrow-14x-x=-2-5\)
\(\Rightarrow-15x=-7\)
\(\Rightarrow x=\frac{7}{15}\)
A
13 tháng 8 2020
a, \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)
\(\Leftrightarrow10x^2-35x+16x-10x^2=5\)
\(\Leftrightarrow-19x=5\Leftrightarrow x=-\frac{5}{19}\)
b, \(x\left(x-\frac{1}{3}\right)-\frac{1}{2}x\left(2x-3\right)=\frac{1}{4}\)
\(\Leftrightarrow x^2-\frac{1}{3}x-x^2+\frac{3}{2}x=\frac{1}{4}\)
\(\Leftrightarrow\frac{7}{6}x=\frac{1}{4}\Leftrightarrow x=\frac{3}{14}\)
c, \(5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)
\(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)
\(\Leftrightarrow-15x+7=0\Leftrightarrow x=\frac{7}{15}\)
Đề thiếu rồi bạn ơi