Cho f(x)= 5x mũ 4 - x mũ 2 (x-3)+3x (x-2)-6x+2
g(x)= 2x mũ 2 (x mũ 2 )-4x mũ 2 + 2 (x + 1) + 5
Tính F(x) + G(x); F(x) - G(X)
Tính a , b để f(x) chia hết cho g(x)
f(x)= x mũ 4 - x mũ 3 + 6x mũ 2 - x + a , g(x)= x mũ 2 - x -2
f( x)= 3 x mũ 3 + 10x mũ 2 - 5 + a , g( x)= 3x + 1
f( x)= x mũ 3 - 3x + a , g(x) =(x - 1) mũ 2
Mình hướng dẫn cách làm chung nhé
f(x) chia hết cho g(x) ⇔ f(x) nhận các nghiệm của g(x) làm nghiệm
Từ đây dễ rồi :]>
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Tính a , b để f(x) chia hết cho g(x)
f(x)= x mũ 4 - x mũ 3 + 6x mũ 2 - x + a , g(x)= x mũ 2 - x -2
f( x)= 3 x mũ 3 + 10x mũ 2 - 5 + a , g( x)= 3x + 1
f( x)= x mũ 3 - 3x + a , g(x) =(x - 1) mũ 2
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Bài 2: Tìm x, biết
a) (x+3) mũ 2 - (x-4)(x+8) = 1
b) (x+3)(x mũ 2 - 3x + 9) -x(x-2)(x+2) = 15
c) (x-2) mũ 2 - (x+3) mũ 2 - 4(x+1) = 5
d) (2x-3)(2x+3) - (x-1) mũ 2 - 3x(x-5) = -44
e) (x-2) mũ 3 - (x-3)(x mũ 2 + 3x + 9) + 6(x+1) mũ 2 = 49
f) 5x(x-3) mũ 2 - 5(x-1) mũ 3 + 15(x+2)(x-2) = 5
g) (x+3) mũ 3 - x(3x+1) mũ 2 + (2x+1)(4x mũ 2 - 2x + 1) - 3x mũ 2 = 42
a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)
\(\Leftrightarrow2x=-40\)
\(\Rightarrow x=-20\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=-12\)
\(\Rightarrow x=-3\)
c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)
\(\Leftrightarrow-14x=14\)
\(\Rightarrow x=-1\)
d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)
\(\Leftrightarrow17x=-34\)
\(\Rightarrow x=-2\)
e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Leftrightarrow24x=24\)
\(\Rightarrow x=1\)
f) \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5\)
\(\Leftrightarrow30x=60\)
\(\Rightarrow x=2\)
g) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=42\)
\(\Leftrightarrow26x=14\)
\(\Rightarrow x=\frac{7}{13}\)
cho mình hỏi đa thức f(X)=-3 mũ 2 +x-1+x mũ 4 -x mũ 3 -x mũ 2 +3 x mũ 4 +2x mũ 3 g(X)=x mũ 4 + x mũ 2 -x mũ 3 + x -5 +5x mũ 3 -x mũ 2 - 3x mũ 4 a thu gon và sắp sếp các đa thúc theo lũy thừa giảm của biến dúp mình bài này với
\(f\left(x\right)=-3x^2+x-1+x^4-x^3-x^2+3x^4+2x^3\)
\(f\left(x\right)=\left(x^4+3x^4\right)-\left(x^3-2x^3\right)-\left(3x^2+x^2\right)+x-1\)
\(f\left(x\right)=4x^4+x^3-4x^2+x-1\)
\(g\left(x\right)=x^4+x^2-x^3+x-5+5x^3-x^2-3x^4\)
\(g\left(x\right)=\left(x^4-3x^4\right)+\left(5x^3-x^3\right)+\left(x^2-x^2\right)+x-5\)
\(g\left(x\right)=-2x^4+4x^3+x-5\)
`@` `\text {Ans}`
`\downarrow`
`a,`
\(f(x) -3x^2 + x - 1 + x^4 - x^3 - x^2 + 3x^4 + 2x^3\)
`= (x^4 +3x^4) + (-x^3 +2x^3) + (-3x^2 - x^2) + x - 1`
`= 4x^4 + x^3 -4x^2 + x -1`
\(g(x) = x^4 + x^2 - x^3 + x - 5 + 5x^3 - x^2 - 3x^4\)
`= (x^4-3x^4) + (-x^3+5x^3) + (x^2 - x^2) + x -5`
`= -2x^4 + 4x^3 +x - 5`
a/ (2x mù -1) mux2 +3(x mũ 2 -2)=0
b/ x mũ 2 - căn 5x =0
c/ 2 x mũ 2 +4x + căn 5 -1=0
d/ x mũ 2 = 2 căn 7x -3
e/ 3x mũ 4 + 2(5x mũ 2 +4)
f/0 x mũ 2 -2 căn 5x +4 =0
g/ 4x mũ 2 +2 căn 13x -3 =0
Bài 1 : Rút gọn
a) (x+2)(x-2) - (x-2)(x+5 )
b) 2x(3x mũ 2 y + 4x mũ 2 y -3)
c) (3x+1) tất cả mũ 2 - (1 -2x) mũ 2
d) x mũ 2 -4-(x+2 ) mũ 2
e) (x-4)(x+4) -2x(x+3) + (x+3) mũ 2
f) (6x+1) mũ 2 -2(6x+1)(6x-1)+(6x-1) mũ 2
(x + 2)(x - 2) - (x - 2)(x + 5)
= (x - 2)(x + 2 - x - 5)
= (x - 2)-3
= -3x + 6
b) 2x(3x2y + 4x2y - 3)
= 2x(7x2y - 3)
= 14x3y - 6x
bài 5 ; tìm x
f, a mũ 2 + x -x mũ 2 - a = a
g, x mũ 2 + 3x - ( 2x + 6 ) = 0
h, 5x + 20 - x mũ 2 - 4x =0
m, x mũ 3 - 5x mũ 2 - x + 5 = 0
n, x ( x - 3 ) - 7x + 21 = 0
Bài 5 :
f, bạn xem lại đề hay là tìm x chứa tham số a ?
g, \(x^2+3x-\left(2x+6\right)=0\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\Leftrightarrow x=-3;x=2\)
h, \(5x+20-x^2-4x=0\Leftrightarrow5\left(x+4\right)-x\left(x+4\right)=0\)
\(\Leftrightarrow\left(5-x\right)\left(x+4\right)=0\Leftrightarrow x=-4;x=5\)
m, \(x^3-5x^2-x+5=0\Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\Leftrightarrow x=\pm1;x=5\)
n, \(x\left(x-3\right)-7x+21=0\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\Leftrightarrow x=3;x=7\)
Bài 1 : Rút gọn
a) (x+2)(x-2) - (x-2)(x+5 )
b) 2x(3x mũ 2 y + 4x mũ 2 y -3)
c) (3x+1) tất cả mũ 2 - (1 -2x) mũ 2
d) x mũ 2 -4-(x+2 ) mũ 2
e) (x-4)(x+4) -2x(x+3) + (x+3) mũ 2
f) (6x+1) mũ 2 -2(6x+1)(6x-1)+(6x-1) mũ 2
a) (x+2)(x-2) - (x-2)(x+5 )
= (x-2) (x+2 - x-5)
= -3 (x-2)
c) \(\left(3x+1\right)^2\) - \(\left(1-2x\right)^2\)
= (3x+1 - 1 +2x) (3x+1 +1-2x)
= 5x (x +2)
d) \(x^2\) - 4 - \(\left(x+2\right)^2\)
= (\(x^2\) - 4 ) - ( x+2) (x+2)
= (x-2) (x+2) - (x+2) (x+2)
= (x+2) (x-2 - x-2)
= -4 (x+2)
e: \(=x^2-16-2x^2-6x+x^2+6x+9=-7\)
b: \(=\left(6x+1-6x+1\right)^2=2^2=4\)
a) 2 mũ 3 : |x-2| =2
b) | 2x + 1 | + | 5x -1 | + 7 mũ 4 = 49
c) | x mũ 3 ( x - 4 )= 5x mũ 2
d) | 2x - 5 | = 2 - x
e) x - | x + 1 | = 4
f) 3x + | 4x + 3 | = 5
a) \(2^3:\left|x-2\right|=2\)
\(\Leftrightarrow8:\left|x-2\right|=2\)
\(\Leftrightarrow\left|x-2\right|=8:2\)
\(\Leftrightarrow\left|x-2\right|=4\)
Xét trường hợp 1: \(x-2=4\)
\(\Rightarrow x=4+2\)
\(\Rightarrow x=6\)
Xét trường hợp 2: \(x-2=-4\)
\(\Rightarrow x=-4+2\)
\(\Rightarrow x=-\left(4-2\right)\)
\(\Rightarrow x=-2\)
Vậy \(x=6\) hoặc \(x=-2\)
b)