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a.
\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(6x^2+21x-2x-7-6x^2+5x-6x+5=16\)
\(\left(6x^2-6x^2\right)+\left(21x-2x+5x-6x\right)-\left(7-5\right)=16\)
\(18x-2=16\)
\(18x=16+2\)
\(18x=18\)
\(x=\frac{18}{18}\)
\(x=1\)
b.
\(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)
\(10x^2+9x-10x^2-15x+2x+3=8\)
\(\left(10x^2-10x^2\right)-\left(15x-9x-2x\right)+3=8\)
\(-4x=8-3\)
\(-4x=5\)
\(x=-\frac{5}{4}\)
c.
\(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)
\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
\(\left(15x^2-15x^2\right)+\left(25x+21x-10x+6x\right)-\left(35+4+2\right)=0\)
\(42x=41\)
\(x=\frac{41}{42}\)
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a, x=-505
b, x=35/8 hoac -37/8
nhung cau con lai thi tong tu
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
a: \(=\dfrac{2x^4+x^3-5x^2-3x-3}{x^2-3}\)
\(=\dfrac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}\)
\(=2x^2+x+1\)
b: \(=\dfrac{x^5+x^2+x^3+1}{x^3+1}=x^2+1\)
c: \(=\dfrac{2x^3-x^2-x+6x^2-3x-3+2x+6}{2x^2-x-1}\)
\(=x+3+\dfrac{2x+6}{2x^2-x-1}\)
d: \(=\dfrac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)
\(=\dfrac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5}{3x^2-2x+1}\)
\(=x^2-2x-5\)
a) \(\text{1- 2x - |3+x| =0 }\)
\(\Leftrightarrow\left|3+x\right|=2x-1\)
\(\Rightarrow\orbr{\begin{cases}3+x=2x-1\\3+x=-\left(2x-1\right)\end{cases}\Rightarrow\orbr{\begin{cases}x-2x=-1-3\\3+x=-2x+1\end{cases}\Rightarrow}\orbr{\begin{cases}-x=-4\\x+2x=1-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\3x=2\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\x=\frac{2}{3}\end{cases}}}\)
\(\text{b) 3x - |2x-1| +2= 2x }\)
\(\Leftrightarrow-\left|2x-1\right|=2x-3x+2\)
\(\Leftrightarrow-\left|2x-1\right|=-x+2\)
\(\Leftrightarrow\left|2x-1\right|=x-2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=x-2\\2x-1=-\left(x-2\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}2x-1=x-2\\2x-1=2-x\end{cases}\Leftrightarrow}\orbr{\begin{cases}2x-x=-2+1\\2x+x=2+1\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\3x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
c) 4/5x + |2/5 - 8/10x| = 1/5
\(\Leftrightarrow\left|\frac{2}{5}-\frac{4}{5}x\right|=\frac{1}{5}-\frac{4}{5}x\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{5}-\frac{4}{5}x=\frac{1}{5}-\frac{4}{5}x\\\frac{2}{5}-\frac{4}{5}x=-\left(\frac{1}{5}-\frac{4}{5}x\right)\end{cases}\Rightarrow\orbr{\begin{cases}-\frac{4}{5}x+\frac{4}{5}x=\frac{1}{5}-\frac{2}{5}\\\frac{2}{5}-\frac{4}{5}x=\frac{4}{5}x-\frac{1}{5}\end{cases}\Leftrightarrow}\orbr{\begin{cases}0=-\frac{1}{5}\left(V\widehat{O}.....LÝ\right)\\-\frac{4}{5}x-\frac{4}{5}x=-\frac{1}{5}-\frac{2}{5}\end{cases}}}\)
\(\Leftrightarrow-\frac{8}{5}x=-\frac{3}{5}\Leftrightarrow x=\left(-\frac{3}{5}\right):\left(-\frac{8}{5}\right)\Leftrightarrow x=\frac{3}{8}\)
Câu hỏi này gửi Nguyễn Ngọc Cát Tường
E carm ơn c nhìu nhìu ạk