8(3x - 2)-10x=2(4-7x)+15 giúp em với
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a. Rút gọn đa thức và sắp xếp theo thứ tự giảm dần của biến..
\(A\left(x\right)=13x^4+3x^2+15x+7x^2-10x^4-7x-6-8x+15\)
\(=\left(13x^4-10x^4\right)+\left(3x^2+7x^2\right)+\left(15x-7x-8x\right)+\left(15-6\right)\)
\(=3x^4+10x^2+9.\)
\(B\left(x\right)=5x^4+10-5x^2-18+3x-10x^2-3x-4x^4\)
\(=\left(5x^4-4x^4\right)+\left(-5x^2-3x^2\right)+\left(3x-3x\right)+\left(10-18\right)\)
\(=x^4-8x^2-8\)
b. Tính M = A(x) + B(x) ; N = A(x) - B(x)
\(M=A\left(x\right)+B\left(x\right)=\left(3x^4+10x^2+9\right)+\left(x^4-8x^2-8\right)\)
\(=\left(3x^4+x^4\right)+\left(10x^2-8x^2\right)+\left(10-8\right)\)
\(=4x^4+2x^2+2\)
\(N=A\left(x\right)-B\left(x\right)=\left(3x^4+10x^2+9\right)-\left(x^4-8x^2-8\right)\)
\(=3x^4+10x^2+9-x^4+8x^2+8\)
\(=\left(3x^4-x^4\right)+\left(10x^2+8x^2\right)+\left(9+8\right)\)
\(=2x^4+18x^2+17\)
\(A=\frac{x^3-3x^2-7x-15}{x^5-x^4-10x^3-38x^2-51x-45}\)
\(=\frac{x^2\left(x-5\right)+2x\left(x-5\right)+3\left(x-5\right)}{x^4\left(x-5\right)+4x^3\left(x-5\right)+10x^2\left(x-5\right)+12x\left(x-5\right)+9\left(x-5\right)}\)
\(=\frac{\left(x-5\right)\left(x^2+2x+3\right)}{\left(x-5\right)\left(x^4+4x^3+10x^2+12x+9\right)}\)
\(=\frac{x^2+2x+3}{x^4+4x^3+10x^2+12x+9}\)
\(=\frac{x^2+2x+3}{\left(x^2\right)^2+2.x^2.2x+\left(2x\right)^2+6x^2+12x+9}\)
\(=\frac{x^2+2x+3}{\left(x^2+2x\right)^2+2.\left(x^2+2x\right).3+3^2}\)
\(=\frac{\left(x^2+2x+3\right)}{\left(x^2+2x+3\right)^2}=\frac{1}{x^2+2x+3}\)
b, \(A=\frac{1}{x^2+2x+3}=\frac{1}{\left(x+1\right)^2+2}\le\frac{1}{2}\forall x\)
Dấu "=" xảy ra khi: \(x+1=0\Rightarrow x=-1\)
Vậy GTLN của A là \(\frac{1}{2}\) khi x = -1
a, 7x + 10x = 5x
17x = 5x
17x - 5x = 0
12x = 0
x =0
2;
a, 4x + 7x = 22
11x = 22
x = 2
b, 12x - 8x = 25
4x = 25
x = \(\dfrac{25}{4}\)
c, \(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)x = \(\dfrac{4}{5}\)
(\(\dfrac{1}{2}-\dfrac{1}{3}\))x = \(\dfrac{4}{5}\)
\(\dfrac{1}{6}\)x = \(\dfrac{4}{5}\)
x = \(\dfrac{4}{5}\) : \(\dfrac{1}{6}\)
x = \(\dfrac{24}{5}\)
a.x2+5x+4
=>x2+4x+x+4
=>x(x+4)+x+4
=>(x+4)(x+1)
b. x2-7x=6
=>x2-x-6x=6
=>x(x-1)-6(x-1)
=>(x-1)(x-6)
c.x2-6x+8
=>x2-2x-4x+8
=>x(x-2)-4(x-2)
=>(x-2)(x-4)
d.x2-10x+21
=>x2-3x-7x+21
=>x(x-3)-7(x-3)
=>(x-3)(x-7)
e.Bạn xem lại í e đi xem có đúng đề không?
g.x2+2x-15
=>x2+5x-3x-15
=>x(x+5)-3(x+5)
=>(x-5)(x-3)
h.x2-3x-28
=>x2+4x-7x-28
=>x(x+4)-7(x+4)
=>(x+4)(x-7)
Nhớ tick cho mình nhé
1, \(x^2+5x+4\)
\(=x^2+x+4x+4\)
\(=x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right).\left(x+4\right)\)
2, \(x^2-7x+6\)
\(=x^2-x-6x+6\)
\(=x\left(x-1\right)-6\left(x-1\right)\)
\(=\left(x+1\right)\left(x-6\right)\)
3, \(x^2-6x+8\)
\(=x^2-2x-4x+8\)
\(=x\left(x-2\right)-4\left(x-2\right)\)
\(=\left(x-2\right)\left(x-4\right)\)
4, \(x^2-10x+21\)
\(=x^2-3x-7x+21\)
\(=x\left(x-3\right)-7\left(x-3\right)\)
\(=\left(x-3\right)\left(x-7\right)\)
5, \(x^2-2x+8\)
\(=x^2+2x-4x+8\)
\(=x\left(x+2\right)-4\left(x-2\right)\)
\(=x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
6, \(x^2+2x-15\)
\(=x^2+5x-3x-15\)
\(=x\left(x+5\right)-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x-3\right)\)
7, \(x^2-3x-28\)
\(=x^2+4x-7x-28\)
\(=x\left(x+4\right)-7\left(x+4\right)\)
\(=\left(x+4\right)\left(x-7\right)\)
\(\dfrac{x}{x^2+x+1}=\dfrac{1}{4}\)
=>\(x^2+x+1=4x\)
=>\(x^2-3x+1=0\)
\(\dfrac{x^5-3x^3-10x+12}{x^4+7x^2+15}\)
\(=\dfrac{x^5-3x^4+x^3+3x^4-9x^3+3x^2+5x^3-15x^2+5x+12x^2-36x+12+21x}{x^4+7x^2+15}\)
\(=\dfrac{x^3\left(x^2-3x+1\right)+3x^2\left(x^2-3x+1\right)+5x\left(x^2-3x+1\right)+12\left(x^2-3x+1\right)+21x}{x^4+7x^2+15}\)
\(=\dfrac{21x}{x^4-3x^3+x^2+3x^3-9x^2+3x+15x^2-45x+15+42x}\)
\(=\dfrac{21x}{x^2\left(x^2-3x+1\right)+3x\left(x^2-3x+1\right)+15\left(x^2-3x+1\right)+42x}\)
\(=\dfrac{21x}{42x}=\dfrac{1}{2}\)
8(3x-2)-10x=2(4-7x)+15
\(\Leftrightarrow24x-16-10x=8-14x+15\)
\(\Leftrightarrow14x-16=-14x+23\)
\(\Leftrightarrow14x+14x=23+16\)
\(\Leftrightarrow28x=39\)
\(\Leftrightarrow x=\frac{39}{28}\)
8(3x-2)-10x = 2(4-7x)+15
24x - 16 - 10x = 8 - 14x + 15
14x - 16 = 23 - 14x
14x + 14x = 23 + 16
28x = 39
x = 39:28
x = 39/28