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\(a,x\left(1-2x\right)-2=\left(2x-3\right)\left(1-x\right)\\ \Leftrightarrow x-2x^2-2=2x-3-2x^2+3x\\ \Leftrightarrow2x-3-2x^2+3x-x+2x^2+2=0\)
\(\Leftrightarrow4x-1=0\\ \Leftrightarrow x=\dfrac{1}{4}\)
\(b,2x\left(x-2\right)+5x-10=0\\ \Leftrightarrow2x\left(x-2\right)+5\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{5}{2}\end{matrix}\right.\)
A = 2x2 - 6xy - 3xy - 6y - 2x2 + 8xy + 6y
= - xy
= \(\frac{2}{3}\)\(x\)\(\frac{3}{4}\)
= \(\frac{1}{2}\)
mk đang bận mấy câu kia tương tự nha
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a: \(\Rightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)
\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)
=>-4x=5
hay x=-5/4
b: \(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
=>42x=41
hay x=41/42
a: =>x^3-3x^2+3x^2-9x+4x-12+a+12 chia hết cho x-3
=>a+12=0
=>a=-12
b: =>2x^2-6x+5x-15+a+15 chia hết cho x-3
=>a+15=0
=>a=-15
c: =>x^3-2x^2-5x^2+20+a-20 chia hết cho x-2
=>a-20=0
=>a=20
e: =>10x^2-15x+8x-12+a+12 chia hết cho 2x-3
=>a+12=0
=>a=-12
f: =>5x^3-x^2+5x^2-x-5x+1-a-1 chia hết cho 5x-1
=>-a-1=0
=>a=-1
Bài 2:
a) x + 2a.(x-y) - y
= 2a.(x-y) + (x-y)
= (x-y).(2a+1)
b) 5a2 - 5ax - 7a + 7x
= 5a.(a-x) - 7.(a-x)
= (a-x).(5a-7)
Bài 1:
a) 5x.(10x+7) - 25x.(2x-3) = 40
50x2 + 35x - 50x2 + 75x = 40
110x = 40
x = 4/11
b) (3x+2).(x-2) - (x-1).(x-3) = 4
3x2 - 6x + 2x - 4 - x2 + 3x + x - 3 = 4
2x2 - 7 = 4
...
bn tự làm tiếp nha
\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)
\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)
b) mình khỏi ghi đề lại ha :3
=> 2x^2 - 4x + 2 + 3x^2 + 12x + 12 - 25x^2 + 1= 15
sau đó bạn gom lại những số vd như là 4x với 12x,..... rồi tính ra đc là
-20x^2 + 8x + 15 = 15
=> -20x^2 + 8x = 0
=> 2x ( -10x + 4 ) = 0
=> 2x = 0 => x= 0
hoặc -10x +4 = 0
=> -10x = -4
=> x = 4/ 10
a) ( 2x-3)^ 2 - ( 2x + 5) ^ 2 = 18
=> 4x^2 - 12x + 9 - ( 4x^2 + 20x + 25 ) = 18
=> 4x^2 - 12x + 9 - 4x^2 - 20x - 25 = 18
=> (4x^2- 4x^2) + (-12x - 20x) + ( 9 -25 ) = 18
=> 0 - 32x - 16 = 18
=> -32x = 32
=> x = -1
bạn đợi mình type câu b :v
a.\(\Leftrightarrow\left(x-1\right)^3+8-x^3+3x\left(x+2\right)=17\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)
\(\Leftrightarrow9x+7=17\)
\(\Leftrightarrow9x=10\Leftrightarrow x=\frac{10}{9}\)
