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Bài giải
a, \(6x\left(4x-3\right)+8x\left(5-3x\right)=43\)
\(24x^2-18x+40x-24x^2=43\)
\(22x=43\)
\(x=\frac{43}{22}\)
b, \(\left(1-7x\right)\left(2x-3\right)-\left(14x-9\right)\left(5-x\right)=30\)
\(2x-14x^2-3+21x-\left(70x-45-14x^2+9x\right)=30\)
\(-56x+42=30\)
\(-56x=-12\)
\(x=\frac{3}{14}\)
a/
\(\Leftrightarrow24x^2-18x+40x-24x^2=43\)
\(\Leftrightarrow22x=43\)
\(\Leftrightarrow x=\frac{43}{22}\)
b/ \(\Leftrightarrow23x-3-14x^2-\left(79x-14x^2-45\right)=30\)
\(\Leftrightarrow-56x=-12\)
\(\Rightarrow x=\frac{3}{14}\)
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)
\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)
\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)
\(=27x^3-4x^2+20x-1\)
b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)
\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)
\(=13x-28x^2-21-x^3\)
c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)
\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)
\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)
\(=16x^2-17+x^3\)
d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)
\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)
\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)
\(=-27x^2+63x-46\)
e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)
\(=12x^2-24x-6x^2-10x-4x^2\)
\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)
\(=2x^2-34x\)
f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)
\(=30x^2-25x-36x+30-3x^2-10x\)
\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)
\(=27x^2-71x+30\)
2) a)\(x\left(x+3\right)-x^2=6\)
\(\Rightarrow x^2+3x-x^2=6\)
\(\Rightarrow\left(x^2-x^2\right)+3x=6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy x=2
b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)
\(\Rightarrow2x^2-10x-2x^2-x=6\)
\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)
\(\Rightarrow-11x=6\)
\(\Rightarrow x=-\dfrac{6}{11}\)
\(\)Vậy \(x=-\dfrac{6}{11}\)
c) x(x+5)-(x+1)(x-2)=7
\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)
\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)
\(\Rightarrow6x=5\)
\(\Rightarrow x=\dfrac{5}{6}\)
Vậy x=\(\dfrac{5}{6}\)
d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)
\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)
\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)
\(\Rightarrow10x-10=10\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy x=2
Tìm x:
1. 3x (2x + 3) - (2x + 5).(3x - 2) = 8
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=0 \)
\(\Leftrightarrow-2x+10=0\Leftrightarrow x=5\)
Vậy x = 5
2. 4x (x -1) - 3(x2 - 5) -x2 = (x - 3) - (x + 4)
\(\Leftrightarrow4x^2-4x-3x^2+15-x^2=x-3-x-4\)
\(\Leftrightarrow-4x+15=-7\)
\(\Leftrightarrow-4x=-22\Leftrightarrow x=\frac{11}{2}\)
Vậy x = \(\frac{11}{2}\)
3. 2 (3x -1) (2x +5) - 6 (2x - 1) (x + 2) = -6
\(\Leftrightarrow2\left(6x^2+15x-2x-5\right)-6\left(2x^2+4x-x-2\right)=-6\)
\(\Leftrightarrow12x^2+30x-4x-10-12x^2-24x+6x+12=-6\)
\(\Leftrightarrow8x=-8\Leftrightarrow x=-1\)
Vậy x = -1
4. 3 ( 2x - 1) (3x - 1) - (2x - 3) (9x - 1) - 3 = -3
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-18x^2+2x+27x-3-3=-3\)
\(\Leftrightarrow18x^2-6x-9x+3-18x^2+2x+27x-6=-3\)
\(\Leftrightarrow14x=0\Leftrightarrow x=0\)
Vậy x = 0
5. (3x - 1) (2x + 7) - ( x + 1) (6x - 5) = (x + 2) - (x - 5)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5=7\)
\(\Leftrightarrow18x=9\Leftrightarrow x=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
6. 3xy (x + y) - (x + y) (x2 + y2 + 2xy) + y3 = 27
\(\Leftrightarrow3x^2y+3xy^2-\left(x+y\right)^3+y^3=27\)
\(\Leftrightarrow3x^2y+3xy^2-x^3-y^3-3x^2y-3xy^2+y^3=27\)
\(\Leftrightarrow-x^3=27\)
\(\Leftrightarrow x=-3\)
Vậy x = -3
7. 3x (8x - 4) - 6x (4x - 3) = 30
\(\Leftrightarrow24x^2-12x-24x^2+12x=30\)
\(\Leftrightarrow0=30\) ( vô lý)
Vậy pt vô nghiệm
8. 3x (5 - 2x) + 2x (3x - 5) = 20
\(\Leftrightarrow15x-6x^2+6x^2-10x=20\)
\(\Leftrightarrow5x=20\Leftrightarrow x=4\)
Vậy x = 4
Bài 1:
\(D=-3x^2+x+15x-5-3\left(2x^2-5x+2\right)\)
\(=-3x^2+16x-5-6x^2+15x-6\)
\(=-9x^2+31x-11\)
\(=-9\cdot\dfrac{1}{9}+\dfrac{31}{3}-11\)
=-11-1+31/3=-12+31/3=-5/3
b: \(E=x^2+x-56-x^2+7x-10=8x-66\)
\(=-\dfrac{8}{5}-66=-\dfrac{338}{5}\)
c: \(F=-3\left(2x^2+x-16x-8\right)-\left(-3x^2+2x-15x+10\right)-4x^2+24x\)
\(=-6x^2+45x+24+3x^2+13x-10-4x^2+24x\)
\(=-4x^2+82x+14\)
\(=-4\cdot9-82\cdot3+14=-268\)
a) Ta có: \(\left(2x+3\right)^2-\left(5+x\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3+5+x\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-3\\3x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{2}\\x=\frac{-8}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-3}{2};\frac{-8}{3}\right\}\)
b) Ta có: \(\left(2x+5\right)^2-\left(2x-5\right)^2=6x+8\)
\(\Leftrightarrow\left(2x+5+2x-5\right)\left(2x+5-2x+5\right)-6x-8=0\)
\(\Leftrightarrow40x-6x-8=0\)
\(\Leftrightarrow34x=8\)
\(\Leftrightarrow x=\frac{8}{34}=\frac{4}{17}\)
Vậy: \(x=\frac{4}{17}\)
c) Ta có: \(\left(4x+3\right)^2=4\left(x-1\right)^2\)
\(\Leftrightarrow16x^2+24x+9=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)
\(\Leftrightarrow12x^2+32x+5=0\)
\(\Leftrightarrow12x^2+2x+30x+5=0\)
\(\Leftrightarrow2x\left(6x+1\right)+5\left(6x+1\right)=0\)
\(\Leftrightarrow\left(6x+1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x+1=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=-1\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)
d) Ta có: \(\left(7x-1\right)\left(3x-2\right)-49x^2+14x=1\)
\(\Leftrightarrow\left(7x-1\right)\left(3x-2\right)-\left(49x^2-14x+1\right)=0\)
\(\Leftrightarrow\left(7x-1\right)\left(3x-2\right)-\left(7x-1\right)^2=0\)
\(\Leftrightarrow\left(7x-1\right)\left[3x-2-\left(7x-1\right)\right]=0\)
\(\Leftrightarrow\left(7x-1\right)\left(3x-2-7x+1\right)=0\)
\(\Leftrightarrow\left(7x-1\right)\left(-4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-1=0\\-4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=1\\-4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{7}\\x=\frac{-1}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{7};\frac{-1}{4}\right\}\)
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha