a) (x+3) ⋮ (x-7)
b) x+8 ⋮ x+2
c) 2x+3 ⋮ x-3 (x>3)
d) 123+x ⋮ 3 và 12≤x≤20
e) 3x+13 ⋮ x-1
f) 12x⋮ 2x+1
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PN
17 tháng 8 2020
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
5 tháng 7 2020
a) \(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)=3\)
\(\Leftrightarrow2x^2+x-x^3-2x^2+x^3-x+3=3\)
\(\Leftrightarrow3=3\)( Luôn đúng với mọi x )
Vậy phương trình nghiệm đúng với mọi x
b) \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x\left(x-1\right)=12x+12\)
\(\Leftrightarrow4x-24-2x^2-3x^3+5x^2-4x+3x^2-3x=12x+12\)
\(\Leftrightarrow-3x^3+6x^2-3x-24=12x+12\)
\(\Leftrightarrow-3x^3+6x^2-3x-24-12x-12=0\)
\(\Leftrightarrow-3x^3+6x^2-15x-36=0\)
Đến đây xem lại đề bạn nhớ :D Tìm thì tìm được nhưng thấy nó sai sai kiểu gì í
c) \(\left(3x+1\right)\left(x-2\right)=\left(2-x\right)\left(-3x-5\right)\)
\(\Leftrightarrow3x\left(x-2\right)+1\left(x-2\right)=2\left(-3x-5\right)-x\left(-3x-5\right)\)
\(\Leftrightarrow3x^2-6x+x-2=-6x-10+3x^2+5x\)
\(\Leftrightarrow3x^2-6x+x+6x-3x^2-5x=-10+2\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\)
d) \(\left(x+3\right)\left(x+5\right)-x\left(x+7\right)=2x+8\)
\(\Leftrightarrow x\left(x+5\right)+3\left(x+5\right)-x\left(x+7\right)=2x+8\)
\(\Leftrightarrow x^2+5x+3x+15-x^2-7x=2x+8\)
\(\Leftrightarrow x^2+5x+3x-x^2-7x-2x=8-15\)
\(\Leftrightarrow-x=-7\)
\(\Leftrightarrow x=7\)
5 tháng 7 2020
a, \(x\left(2x-1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)=3\)
\(\Leftrightarrow2x^2-x-x^3-2x^2+x^3-x+3=3\)
\(\Leftrightarrow-2x=0\Leftrightarrow x=0\)
b, \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x\left(x-1\right)=12x+12\)
\(\Leftrightarrow4x-24-2x^2-3x^3+5x^2-4x+3x^2-3x=12x+12\)
\(\Leftrightarrow-3x-24+6x^2-3x^3=12x+12\)
\(\Leftrightarrow-15x-36+6x^2-3x^3=0\)
Lớp 8 chưa hc vô tỉ đâu ... vô nghiệm
c, \(\left(3x+1\right)\left(x-2\right)=\left(2-x\right)\left(-3x-5\right)\)
\(\Leftrightarrow3x^2-5x-2=-x-10+3x^2\)
\(\Leftrightarrow-4x+8=0\Leftrightarrow x=2\)
d, \(\left(x+3\right)\left(x+5\right)-x\left(x+7\right)=2x+8\)
\(\Leftrightarrow x^2+8x+15-x^2-7x=2x+8\)
\(\Leftrightarrow x+15=2x+8\Leftrightarrow-x+7=0\Leftrightarrow x=7\)
9 tháng 11 2021
Bài 2:
c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
8 tháng 9 2021
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
8 tháng 9 2021
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
VV
7 tháng 1 2022
\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)
9 tháng 9 2021
\(d,=24x^2-38x+3\\ e,=x^2-12x+35\\ f,=\left(x^2-144\right)\left(4x-1\right)=4x^3-x^2-576x+144\)