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\(\left(x^2+x+1\right)\left(x^2+x+5\right)-21=x^4+x^3+5x^2+x^3+x^2+5x+x^2+x+5-21=x^4+2x^3+7x^2+6x-16=\left(x-1\right)\left(x+2\right)\left(x^2+x+8\right)\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+4\right)-21\)
\(=\left(x^2+x+1\right)^2+4\left(x^2+x+1\right)-21\)
\(=\left(x^2+x+1\right)^2-3\left(x^2+x+1\right)+7\left(x^2+x+1\right)-21\)
\(=\left(x^2+x+1\right)\left(x^2+x-2\right)+7\left(x^2+x-2\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+8\right)\)
\(=\left(x-1\right)\left(x-2\right)\left(x^2+x+8\right)\)
5 x 2 + 10xy – 4x – 8y = (5 x 2 + 10xy) – (4x + 8y)
= 5x(x + 2y) – 4(x + 2y) = (5x – 4)(x + 2y)
Đáp án cần chọn là: C
5 x 2 + 10 x y – 4 x – 8 y = 5 x 2 + 10 x y – 4 x + 8 y = 5 x x + 2 y – 4 x + 2 y = 5 x - 4 x + 2 y
Đáp án cần chọn là: C
Bài 1:
$5x+10=5(x+2)$
Bài 2:
Tại $x=8$ thì $x^2+4x+4=(x+2)^2=(8+2)^2=10^2=100$
Bài 3:
$x^2-6x+9=x^2-2.3.x+3^2=(x-3)^2$
Bài 4:
Diện tích mảnh đất là:
$(x+5)(x-5)=24$
$\Leftrightarrow x^2-25=24$
$\Leftrightarrow x^2=49$
$\Rightarrow x=7$ (do $x>5$)
Chiều dài mảnh đất là: $x+5=7+5=12$ (m)
\(a,=3\left(x-5\right)-x\left(x-5\right)=\left(3-x\right)\left(x-5\right)\\ b,=7\left(x^2-2xy+y^2\right)=7\left(x-y\right)^2\\ c,=\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)=\left(x-y\right)^2\left(x+y\right)^2\\ d,=\left(y^2-6y+9\right)-25x^2=\left(y-3\right)^2-25x^2=\left(y-5x-3\right)\left(y+5x-3\right)\)
`
` x^2 – 3x + 2`
`= x^2 – x – 2x + 2` (Tách `–3x = – x – 2x)`
`= (x^2 – x) – (2x – 2)`
\(\Leftrightarrow x^3-2x^2+x^2-2x+x-2\)
\(\Leftrightarrow x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+x+1\right)\)
Ta có:
\(\left(x-1\right)\left(x+2\right)\left(x-3\right)\left(x+4\right)=144\)
\(\Leftrightarrow\left(x^2-x-2\right)\left(x^2-x-12\right)=144\)
Đặt \(x^2-x-7=m\left(1\right)\),Ta có:
\(\Leftrightarrow\left(m+5\right)\left(m-5\right)=144\)
\(\Leftrightarrow m^2=169\Rightarrow m=13\)
Thay \(\left(1\right)=13\)
\(\Rightarrow x^2-x-7=13\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)
\(\Rightarrow x=5;x=-4\)
-Đặt \(t=\left(x^2-x+1\right)\)
\(\left(x^2-x+1\right)^2-5x\left(x^2-x+1\right)+4x^2\)
\(=t^2-5xt+4x^2\)
\(=t^2-4xt-xt+4x^2\)
\(=t\left(t-4x\right)-x\left(t-4x\right)\)
\(=\left(t-4x\right)\left(t-x\right)\)
\(=\left(x^2-x+1-4x\right)\left(x^2-x+1-x\right)\)
\(=\left(x^2-5x+1\right)\left(x^2-2x +1\right)\)
\(=\left(x^2-5x+1\right)\left(x-1\right)^2\)
= x4 - 10x2 + 25 + 144
= x4 - 10x2 + 179