7x-(5x-12)=-34
\(\frac{7x-12}{5x-2}=\frac{9-7x}{-5x+1}\)
\(\frac{7x-12}{5x-2}=\frac{9-7x}{-5x+1}\)
\(=>\frac{2x-10}{5x-2}+1=\frac{-7x+9}{-5x+1}=\frac{-2x+8}{-5x+1}+1\)
\(=>\frac{2x-10}{5x-2}=\frac{-2x+8}{-5x+1}=\frac{-\left(2x-8\right)}{-\left(5x-1\right)}=\frac{2x-8}{5x-1}\)
\(=>\frac{2x-10}{5x-2}=\frac{2x-8}{5x-1}\)
\(=>\frac{5x-1}{5x-2}=\frac{2x-8}{2x-10}\)
\(=>\frac{-1}{5x-2}=\frac{-2}{2x-10}\)
\(=>-2x+10=-10x+4\)
\(=>-2x+10x=4-10=-6\)
\(=>8x=-6\)
\(=>x=\frac{-6}{8}=\frac{-3}{4}\)
P/s ko chắc :v
Ta có: \(\frac{7x-12}{5x-2}=\frac{7x-9}{5x-1}\)
\(\Rightarrow\left(7x-12\right).\left(5x-1\right)=\left(7x-9\right).\left(5x-2\right)\)
\(\Leftrightarrow35x^2-7x-60x+12=35x^2-14x-45x+18\)
\(\Leftrightarrow\left(35x^2-35x^2\right)-\left(7x+60x-14x-45x\right)=18-12\)
\(\Leftrightarrow8x=6\)
\(\Leftrightarrow x=\frac{3}{4}\)
Vậy \(\Leftrightarrow x=\frac{3}{4}\)
\(\frac{7x-12}{5x-2}=\frac{9-7x}{-5x+1}\)
\(\left(7x-12\right)\left(-5x+1\right)=\left(9-7x\right)\left(5x-2\right)\)
\(-35x^2+7x+60x-12=45x-18-35x^2+14x\)
\(67x-12=59x-18\)
\(67x-12-59x=18\)
\(8x-12=-18\)
\(8x=-18+12\)
\(8x=-6\)
\(x=-\frac{6}{8}=-\frac{3}{4}\)
Bài 5: Tìm a , b để các đa thức sau:
1) x^4+6x^3+7x^2-6x+a chia hết cho x2+3x-1
2) x^4-x^3+6x^2-x+a chia hết cho x^2- x+5
3) x^3+3x^2+5x+a chia hết cho x+3
4) x^3+2x^2-7x+a chia hết cho 3x -1
5) 2x^2+ax+1 chia cho x-3 dư 4
3: \(\Leftrightarrow a-15=0\)
hay a=15
tính:
(-7)-(-12)+(-5)
(-2)x3 mũ 2+7x(-4)-(-34)
12x(-82)+(-12)x18
*) (-7) - (-12) + (-5)
= [(-7) + (-5)] + 12
= (-12) + 12
= 0
*) (-2) . 32 + 7 . (-4) - (-34)
= (-2) . 9 + (-28) + 34
= -18 + (-28) + 34
= -46 + 34
= -12
*) 12 . (-82) + (-12).18
= (-12) . 82 + (-12) . 18
= (-12) . (82 + 18)
= (-12) . 100
= -1200
( - 7 )- ( - 12 ) + ( - 5) = 0
( - 2 ) x 32+ 7 . (-4) - ( -34) = - 970
12 x ( -82 ) + (-12) x18 = - 1200
( - 7 )- ( - 12 ) + ( - 5)
= (-7) + 12 + (-5)
= 5 + (-5)
= 0
(- 2) x 32+ 7 . (-4) - ( -34)
= (-2) x 9 + 7 x (-4) - (-34)
= -18 + (-28) - (-34)
= -18 + (-28) + 34
= -46 + 34
= -12
12 x ( -82 ) + (-12) x18
= -984 + (-216)
= -1200
a) 5x^2-7x-12=0
a) \(5x^2-7x-12=0\)
Ta có: \(a=5;b=-7;c=-12\)
\(\Rightarrow\Delta=b^2-4ac=\left(-7\right)^2-4.5.\left(-12\right)=289\)
Vậy phương trình có 2 nghiệm:
\(\left\{{}\begin{matrix}x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-\left(-7\right)-\sqrt{289}}{2.5}=-1\\x_2=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-\left(-7\right)+\sqrt{289}}{2.5}=2,4\end{matrix}\right.\)
Phân tích đa thức thành nhân tử ( bằng kĩ thuật bổ sung hằng đẳng thức ):
a.x^2 - 5x + 6
b.x^2 + 5x + 6
c.x^2 - 7x +12
d.x^2 + 7x +12
a)x2-5x+6=(x2-2x)-(3x-6)=x(x-2)-3(x-2)(=(x-2)(x-3)
b)x2+5x+6=(x2+2x)+(3x+6)=x(x+2)+3(x+2)=(x+2)(x+3)
c)x2-7x+12=(x2-3x)-(4x-12)=x(x-3)-4(x-3)=(x-3)(x-4)
d)x2+7x+12=(x2+3x)+(4x+12)=x(x+3)+4(x+3)=(x+3)(x+4)
Tính: a,x²-5x+6/ x²+7x+12 : x²-4x+4/ x²+3x b,x²+2x-3/ x²+3x-10 : x²+7x+12/ x²-9x+14
a: \(\dfrac{x^2-5x+6}{x^2+7x+12}:\dfrac{x^2-4x+4}{x^2+3x}\)
\(=\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x+3\right)\left(x+4\right)}\cdot\dfrac{x\left(x+3\right)}{\left(x-2\right)^2}\)
\(=\dfrac{x\left(x-3\right)}{\left(x-2\right)\left(x+4\right)}\)
b: \(\dfrac{x^2+2x-3}{x^2+3x-10}:\dfrac{x^2+7x+12}{x^2-9x+14}\)
\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x+5\right)\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x-7\right)}{\left(x+3\right)\left(x+4\right)}\)
\(=\dfrac{\left(x-1\right)\left(x-7\right)}{\left(x+5\right)\left(x+4\right)}\)
1) x2 -7x + 10 = x2 - 2x - 5x + 10 = x(x - 2) - 5(x - 2) = (x - 5)(x - 2)
2) x2 + 3x + 2 = x2 + 2x + x + 2 = x(x + 2) + (x + 2) = (x + 1)(x + 2)
3) x2 - 7x + 12 = x2 - 3x - 4x + 12 = x(x - 3) - 4(x - 3) = (x - 3)(x - 4)
4) x2 + 7x + 12 = x2 + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4)
5) 16x - 5x2 - 3 = 15x - 5x2 + x - 3 = -5x(x - 3) + (x - 3) = (x - 3)(1 - 5x)
6) 6x2 + 7x - 3 = 6x2 - 2x + 9x - 3 = 2x(3x - 1) + 3(3x - 1) = (2x + 3)(3x - 1)
7) 3x2 - 3x - 6 = 3x2 - 6x + 3x - 6 = 3x(x - 2) + 3(x - 2) = (x - 2)(3x + 3) = 3(x - 2)(x + 1)
8) 3x2 + 3x - 6 = 3x2 - 3x + 6x - 6 = 3x(x - 1) + 6(x - 1) = (x - 1)(3x + 6) = 3(x - 1)(x + 2)
9) 6x2 - 13x + 6 = 6x2 - 9x - 4x + 6 = 3x(2x - 3) - 2(2x - 3) = (3x - 2)(2x - 3)
10) 6x2 + 15x + 6 = 6x2 + 12x + 3x + 6 = 6x(x + 2) + 3(x + 2) = (x + 2)(6x + 3) = 3(x + 2)(3x + 1)
11) 6x2 - 20x + 6 = 6x2 - 18x - 2x + 6 = 6x(x -3) - 2(x - 3) = (6x - 2)(x - 3) = 2(3x - 1)(x - 3)
12) 8x2 + 5x - 3 = 8x2 + 8x - 3x - 3 = 8x(x + 1) - 3(x + 1) = (x + 1)(8x - 3)
[7x-19]-[5x-12]=3x-[9-2x]-1
\(\left(7x-19\right)-\left(5x-12\right)=3x-\left(9-2x\right)-1\)
\(\Leftrightarrow7x-19-5x+12=3x-9+2x-1\)
\(\Leftrightarrow-3x=3\)
\(\Leftrightarrow x=-1\)
bài 2: cho đa thức
A(x)=5/6x mũ3 - 12/7x mũ2+5x+5/7x mux2 +1/6x mux3 - 3x+9