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\(A_3=\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)
Đặt \(t=x^2+4x+10\)
\(A_3=t^2-7\left(t^2+1\right)+7\)
\(=-6t^2\)
Thay vào : \(-6\left(x^2+4x+10\right)^2\)
2 , \(A_1=\left(t^2+3x\right)^2-2\left(x^2+3x\right)-8\)
Đặt \(t=x^2-3x\)
\(A_1=t^2-2x-8=\left(t-4\right)\left(t+2\right)\)
\(=\left(x^2+3x+2\right)\left(x^2+3x-4\right)\)
1.
7x(2x-1)=14x2-7x
2
a. x2+2x=x(x+2)
b.x2-xy+3x-3y
=x(x-y)+3(x-y)
=(x+3)(x-y)
Câu 2:
1. 2x/2x-5 - 5/2x-5
=2x-5/2x-5
=1
2. (6x3-7x2-x+2) : (x-1)=6x2-x-2
b1:
câu a,f áp dụng a2-b2=(a-b)(a+b)
câu b,c áp dụng a3-b3=(a-b)(a2+ab+b2)
câu d: \(x^2+2xy+x+2y=x\left(x+2y\right)+\left(x+2y\right)=\left(x+1\right)\left(x+2y\right)\)
câu e: \(7x^2-7xy-5x+5y=7x\left(x-y\right)-5\left(x-y\right)=\left(7x-5\right)\left(x-y\right)\)
câu g xem lại đề
\(x^2+\dfrac{1}{2}x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)
1.
= 4x\(^{^{ }2}\)-4x-9x+9
=4x(x-1)-9(x-1)
=(4x-9)(x-1)
a) A = (x2 - 2.x.3 + 32) - (3y)2
A = (x - 3)2 - (3y)2
A = (x - 3 - 3y)(x-3+3y)
b) B = (x-1)3 + 2(x-1)(x+1)
B=(x-1)[(x-1)2 - 2(x+1)]
B = (x-1)[x2 - 4x - 3]
a: \(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15x^2-60-5=0\)
\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-65=0\)
\(\Leftrightarrow30x-60=0\)
hay x=2
b: \(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2=42\)
\(\Leftrightarrow9x^3+6x^2+27x+28-9x^3-6x^2-x=42\)
=>26x=14
hay x=7/13
a) \(A_4=\left(x^2-3x+5\right)^2+7x\cdot\left(x^2-3x+5\right)+12x^2\)
\(=\left(x^2-3x+5\right)^2+4x\cdot\left(x^2-3x+5\right)+3x\left(x^2-3x+5\right)+12x^2\)
\(=\left(x^2-3x+5\right)\left(x^2-3x+5+4x\right)+3x\left(x^2-3x+5+4x\right)\)
\(=\left[\left(x^2-3x+5\right)+3x\right]\cdot\left(x^2-3x+5+4x\right)\)
\(=\left(x^2-3x+5+3x\right)\left(x^2+x+5\right)\)
\(=\left(x^2+5\right)\left(x^2+x+5\right)\)
\(A_5=2\left(x^2+5x-2\right)^2-7\left(x^2+5x-2\right)\left(x^3+3\right)+5\left(x^2+3\right)^2\)
Đặt \(x^2+5x-2=a;x^3+3=b\),Ta có:
\(2a^2-7ab+5b^2=2a^2-5ab-2ab+5b^2=a\left(2a-5b\right)-b\left(2a-5b\right)=\left(2a+5b\right)\left(a-b\right)\)
Thay \(x^2+5x-2=a;x^3+3=b\),ta có:
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bn làm nốt nhé