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Ta có : x3 - 7x + 6
= x3 - x - 6x + 6
= x(x2 - 1) - 6(x - 1)
= x(x + 1)(x - 1) - 6(x - 1)
= (x - 1) [x(x + 1) - 6]
= (x - 1) (x2 + x - 6) .
CÁC Ý SAU TƯƠNG TỰ
1
x3-7x+6
=x3+0x2-7x +6
= x3-x2+x2-x-6x+6
=(x3-x2)+(x2-x)-(6x-6)
=x2(x-1)+x(x-1)-6(x-1)
=(x-1)(x2+x-6)
=(x-1)(x2+3x-2x-6)
=(x-1)[x(x+3)-2(x+3)]
=(x-1)(x-2)(x+3)
7) (x+2)(x+3)(x+4)(x+5)-24
=(x+2)(x+5) (x+3)(x+4)-24
=[x(x+5)+2(x+5)][x(x+4)+3(x+4)]-24
=[x2+5x+2x+10][x2+4x+3x+12]-24
=[x2+7x+10][x2+7x+12]-24
đặt a=x2+7x+10
=>x2+7x+12=a+2
=a(a+2)-24
=a2+2a-24
=a2+6a-4a-24
=(a2+6a)-(4a+24)
=a(a+6)-4(a+6)
=(a+6)(a-4)
thay a= x2+7x+10 vào ta được
(x2+7x+10+6)(x2+7x+10-4)
=(x2+7x+16)(x2+7x+6)
giai phuong trinh
1, (x-2)(x-1)(x-8)(x-4)=4x^2
2, (x^2+5x+6)(x^2+20x+96)=4x^2
3, 3(x^2+2x-1)^2-2(x^2+3x-1)^2+5x^2=0
a) \(\left(x+8\right)\left(x+6\right)=104+x^2\Leftrightarrow x^2+6x+8x+48=104+x^2\)
\(\Leftrightarrow x^2+6x+8x-x^2=104-48\Leftrightarrow14x=56\Leftrightarrow x=\dfrac{56}{14}=4\)
vậy \(x=4\)
b) \(\left(x+1\right)\left(x+2\right)-\left(x-3\right)\left(x+4\right)=6\)
\(\Leftrightarrow x^2+2x+x+2-\left(x^2+4x-3x-12\right)=6\)
\(\Leftrightarrow x^2+2x+x+2-x^2-4x+3x+12=6\)
\(\Leftrightarrow2x+14=6\Leftrightarrow2x=6-14=-8\Leftrightarrow x=\dfrac{-8}{2}=-4\)
vậy \(x=-4\)
c) \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(\Leftrightarrow4x^2-20x-\left(4x^2-3x-4x+3\right)=5\)
\(\Leftrightarrow4x^2-20x-4x^2+3x+4x-3=5\)
\(\Leftrightarrow-13x-3=5\Leftrightarrow-13x=5+3=8\Leftrightarrow x=\dfrac{8}{-13}=\dfrac{-8}{13}\)
vậy \(x=\dfrac{-8}{13}\)
d) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)
\(\Leftrightarrow3x^2-6x-4x+8=3x^2-27x-3\)
\(\Leftrightarrow3x^2-6x-4x-3x^2+27x=-3-8\)
\(\Leftrightarrow17x=-11\Leftrightarrow x=\dfrac{-11}{17}\) vậy \(x=\dfrac{-11}{17}\)
e) câu này đề bị thiếu rồi nha bn
f) \(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\)
\(\Leftrightarrow5x^2-15x=5x^2-x-10x+2-5\)
\(\Leftrightarrow5x^2-15x-5x^2+x+10x=2-5\)
\(\Leftrightarrow-4x=-3\Leftrightarrow x=\dfrac{-3}{-4}=\dfrac{3}{4}\) vậy \(x=\dfrac{3}{4}\)
a) \(\left(x+8\right)\left(x+6\right)=104+x^2\)
\(\Leftrightarrow x^2+14x+48=104+x^2\)
\(\Leftrightarrow14x=56\)
\(\Rightarrow x=4\)
b) \(\left(x+1\right)\left(x+2\right)-\left(x-3\right)\left(x+4\right)=6\)
\(\Leftrightarrow x^2+3x+2-x^2-7x+12=6\)
\(\Leftrightarrow-4x=-8\)
\(\Rightarrow x=2\)
c) \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(\Leftrightarrow4x^2-20x-4x^2+3x+4x-3=5\)
\(\Leftrightarrow-13x=8\)
\(\Rightarrow x=\dfrac{-8}{13}\)
d) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)
\(\Leftrightarrow3x^2-10x+8=3x^2-27x-3\)
\(\Leftrightarrow17x=-11\)
\(\Rightarrow x=\dfrac{-11}{17}\)
e) \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)
\(\Leftrightarrow x^2-9x+20-x^2+x+2=7\)
\(\Leftrightarrow-8x=-15\)
\(\Rightarrow x=\dfrac{15}{8}\)
f) \(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\)
\(\Leftrightarrow5x^2-15x=5x^2-11x+2-5\)
\(\Leftrightarrow-4x=-3\)
\(\Rightarrow x=\dfrac{3}{4}\)
a) Ta có: \(4x^2-6x\)
\(=2x\left(2x-3\right)\)
b) Ta có: \(9x^4y^3+3x^2y^4\)
\(=3x^2y^3\left(3x^2+y\right)\)
c) Ta có: 3(x-y)-5x(y-x)
=3(x-y)+5x(x-y)
=(x-y)(3+5x)
d) Ta có: \(x^3-2x^2+5x\)
\(=x\left(x^2-2x+5\right)\)
e) Ta có: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=\left(x+3y\right)\left(5-15x\right)\)
\(=5\left(x+3y\right)\cdot\left(1-3x\right)\)
f) Ta có: \(2x^2\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^2+4\right)\)
\(=2\left(x+1\right)\left(x^2+2\right)\)
Nhiều vậy ai làm hết được :P
1) \(\frac{3x-2}{3}-2=\frac{4x+1}{4}\)
\(\Leftrightarrow\frac{3x-8}{3}=\frac{4x-1}{4}\)
\(\Leftrightarrow4\left(3x-8\right)=3\left(4x-1\right)\)
\(\Leftrightarrow12x-32=12x-3\)(vô lí)
Vậy pt vô nghiệm
P/s: mấy câu sau tương tự thôi mà :)))
nhăm nhe 1 câu thôi
\(10,\frac{3+5x}{5}-3=\frac{9x-3}{4}\)
\(\Leftrightarrow\frac{3+5x-15}{5}=\frac{9x-3}{4}\)
\(\Leftrightarrow\frac{-12+5x}{5}=\frac{9x-3}{4}\)
\(\Leftrightarrow\left(-12+5x\right)5=\left(9x-3\right)4\)
\(\Leftrightarrow-60+25x=36x-12\)
\(\Leftrightarrow26x-36x=-12+60\)
\(\Leftrightarrow-10x=48\)
\(\Leftrightarrow x=-4,8\)