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\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
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) 5/4+x=2/3
B) -x-2=5/4
C)4x+1/3=3/2
Đ) 1/3-2/5+3x=3/4
E) 3x+7+2x=4x-3
G) 3x(2x-3)-2x(3x-4)=15
H) x^2-x=0
a) \(x=-\frac{7}{12}\)
b) \(x=-\frac{13}{4}\)
c) \(x=\frac{7}{24}\)
d) \(x=\frac{49}{180}\)
e) \(x=-10\)
g) \(x=15\)
h) \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
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: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
Bạn lưu ý ghi rõ yêu cầu đề. Ghi đầy đủ thì mọi người mới biết đường hỗ trợ bạn chứ?
\(2,\left(3x+1\right)\left(2x-3\right)-\left(3x-4\right)^2-2x\left(3x+1\right)\)
\(=6x^2-9x+2x-3-9x^2+24x-16-6x^2-2x\)
\(=\left(6x^2-9x^2-6x^2\right)+\left(-9x+2x+24x-2x\right)+\left(-3-16\right)\)
\(=-9x^2+15x-19\)
\(3,\left(3x+2\right)^2-\left(4x+1\right)\left(2x-3\right)-2x\left(3x+1\right)\)
\(=9x^2+12x+4-8x^2+12x-2x+3-6x^2-2x\)
\(=\left(9x^2-8x^2-6x^2\right)+\left(12x+12x-2x-2x\right)+\left(4+3\right)\)
\(=-5x^2+20x+7\)
\(4,\left(3x-2\right)^2-\left(3x-2\right)\left(x+1\right)-4x\left(2x+3\right)\)
\(=9x^2-12x+4-\left(3x^2+3x-2x-2\right)-8x^2-12x\)
\(=9x^2-12x+4-3x^2-3x+2x+2-8x^2-12x\)
\(=-2x^2-25x+6\)
\(5,\left(2x-5\right)^2-4x\left(x-3\right)-\left(3x+1\right)\left(2x-5\right)\)
\(=4x^2-20x+25-4x^2+12x-\left(6x^2-15x+2x-5\right)\)
\(=4x^2-20x+25-4x^2+12x-6x^2+15-2x+5\)
\(=-6x^2+5x+30\)