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a) = 12x2 -12x2 +20x -10x +17 =0
10x = -17
x = -17/10
a) 4x2 - 2x + 3 - 4x.(x - 5) = 7x - 3
--> 4x2 - 2x + 3 - 4x2 + 20x = 7x - 3
--> 4x2 - 2x - 4x2 + 20x - 7x = -3 - 3
--> 11x = -6
--> x = \(\frac{-6}{11}\)
b) -3x.(x - 5) + 5.(x - 1) + 3x2 = 4x
--> -3x2 + 15x + 5x - 5 + 3x2 = 4x
--> -3x2 + 15x + 5x + 3x2 - 4x = 5
--> 16x = 5
--> x = \(\frac{5}{16}\)
c) 7x.(x - 2) - 5.(x - 1) = 21x2 - 14x2 + 3
--> 7x2 - 14x - 5x + 5 = 7x2 + 3
--> 7x2 - 14x - 5x - 7x2 = -5 + 3
--> -19x = -2
--> x = \(\frac{2}{19}\)
d) 3.(5x - 1) - x.(x - 2) + x2 - 13x = 7
--> 15x - 3 - x2 + 2x + x2 - 13x = 7
--> 15x - x2 + 2x + x2 - 13x = 3 + 7
--> 4x = 10
--> x = \(\frac{5}{2}\)
e) \(\frac{1}{5}\)x.(10x - 15) - 2x.(x - 5) = 12
--> 2x2 - 3x - 2x2 + 10x = 12
--> 7x = 12
--> x = \(\frac{12}{7}\)
~ Học tốt ~
a) 4x2 - 2x + 3 - 4x(x - 5) = 7x - 3
=> 4x2 - 2x + 3 - 4x2 + 20x = 7x - 3
=> 18x + 3 = 7x - 3
=> 18x - 7x = -3 - 3
=> 11x = -6
=> x = -6/11
b) -3x(x - 5) + 5(x - 1) + 3x2 = 4x
=> -3x2 + 15x + 5x - 5 + 3x2 = 4x
=> 20x - 5 = 4x
=> 20x - 4x = 5
=> 16x = 5
=> x = 5/16
\(c,7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)
\(\Leftrightarrow7x^2-14x-5x+5=7x^2+3\)
\(\Leftrightarrow7x^2-7x^2-19x=3-5\)
\(\Leftrightarrow-19x=-2\)
\(\Leftrightarrow x=\frac{2}{19}\)
a) \(3x\left(x-3\right)-5x\left(x+7\right)\)
\(=3x^2-9x-5x^2-35x\)
\(=-2x^2-44x\)
b) \(\dfrac{1}{5}x\left(10x-15\right)-2x\left(x-5\right)-12\)
\(=2x^2-3x-2x^2+10x-12\)
\(=7x-12\)
1
\(-3x\left(x-5\right)+5\left(x-1\right)+3x^2=4-x\)
=> \(-3x^2+15x+5x-5+3x^2=4-x\)
=> \(20x-5=4-x\)
=> \(21x=9\)
=> \(x=\dfrac{3}{7}\)
Vậy x = \(\dfrac{3}{7}\)
2,
\(7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)
=> \(7x^2-14x-5x+5=7x^2+3\)
=> \(-14x-5x+5=3\)
=> \(-19x=-2\)
=> \(x=\dfrac{2}{19}\)
Vậy \(x=\dfrac{2}{19}\)
3,
\(3\left(5x-1\right)-x\left(x-2\right)+x^2-13x=7\)
=> \(15x-3-x^2+2x+x^2-13x=7\)
=> \(4x-3=7\)
=> 4x = 10
=> x = \(\dfrac{5}{2}\)
Vậy x = \(\dfrac{5}{2}\)
4,
\(\dfrac{1}{5}x\left(10x-15\right)-2x\left(x-5\right)=12\)
=> \(2x^2-3x-2x^2+10x=12\)
=> 7x = 12
=> x = \(\dfrac{12}{7}\)
Vậy x = \(\dfrac{12}{7}\)
a) Ta có: \(P\left(x\right)=x^7-80x^6+80x^5-80x^4+...+80x+15\)
\(=x^7-x^6\left(x+1\right)+x^5\left(x+1\right)-...+x\left(x+1\right)+15\)
\(=x^7-x^7-x^6+x^6+x^5-...+x^2+x+15\)
\(=x+15\)
Thay x=79 vào biểu thức \(P\left(x\right)=x+15\), ta được:
\(P\left(79\right)=79+15=94\)
11) \(3x\left(x-1\right)+5\left(1-x\right)=\left(3x-5\right)\left(x-1\right)\)
12) \(2\left(2x-1\right)+3\left(1-2x\right)=1-2x\)
13) \(10x\left(x-y\right)-8y\left(y-x\right)=2\left(x-y\right)\left(5x+4y\right)\)
14) \(3x\left(y+2\right)-3\left(y+2\right)=3\left(x-1\right)\left(y+2\right)\)
15) \(x^2-y^2-2x+2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
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
a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)
\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)
\(< =>12-2+4x-2x^2=6x^2-13x+6\)
\(< =>10+4x-2x^2-6x^2+13x-6=0\)
\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)
b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)
\(< =>x-9=0< =>x=9\)
c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)
\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)
d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)
\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)
e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)
\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)
f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)
\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)
g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)
\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)
h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
\(< =>x^2-16-6x+4=x^2-8x+16\)
\(< =>x^2-6x-12-x^2+8x-16=0\)
\(< =>2x-28=0< =>x=\frac{28}{2}=14\)
q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề
\(\frac{1}{5}\left(10x-15\right)-2x\left(x-5\right)=12\)
\(\Leftrightarrow2x^2-3x-2x^2+10x=12\)
\(\Leftrightarrow-3x+10x=12\)
\(\Leftrightarrow7x=12\)
\(\Leftrightarrow x=\frac{12}{7}\)