Mọi người giúp e câu 6 với
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1: Ta có: \(A=25x^4-24x^2-1\)
\(=25x^4-25x^2+x^2-1\)
\(=\left(x^2-1\right)\left(25x^2+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(25x^2+1\right)\)
2: Ta có: \(A=64x^4+63x^2-1\)
\(=64x^4+64x^2-x^2-1\)
\(=\left(x^2+1\right)\left(64x^2-1\right)\)
\(=\left(x^2+1\right)\left(8x-1\right)\left(8x+1\right)\)
3: Ta có: \(A=x^4-15x^2+50\)
\(=x^4-5x^2-10x^2+50\)
\(=\left(x^2-5\right)\left(x^2-10\right)\)
4: Ta có: \(A=-10x^4+9x^2+1\)
\(=-10x^4+10x^2-x^2+1\)
\(=\left(x^2-1\right)\left(-10x^2-1\right)\)
\(=-\left(10x^2+1\right)\left(x-1\right)\left(x+1\right)\)
Lời giải:
Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:
$|x+2020|+|x+2021|=|x+2020|+|-(x+2021)|$
$\geq |x+2020-(x+2021)|=1$
Vậy GTNN của biểu thức là $1$. Giá trị này đạt tại $(x+2020).-(x+2021)\geq 0$
$(x+2020)(x+2021)\leq 0$
$-2021\leq x\leq -2020$
37: Ta có: \(A=8x^2-2x-3\)
\(=8x^2-6x+4x-3\)
\(=2x\left(4x-3\right)+\left(4x-3\right)\)
\(=\left(4x-3\right)\left(2x+1\right)\)
38: Ta có: \(A=8x^2+2x-3\)
\(=8x^2+6x-4x-3\)
\(=2x\left(4x+3\right)-\left(4x+3\right)\)
\(=\left(4x+3\right)\left(2x-1\right)\)
39: Ta có: \(A=-8x^2+5x+3\)
\(=-8x^2+8x-3x+3\)
\(=-8x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x-1\right)\left(-8x-3\right)\)
40: Ta có: \(A=8x^2-10x-3\)
\(=8x^2-12x+2x-3\)
\(=4x\left(2x-3\right)+\left(2x-3\right)\)
\(=\left(2x-3\right)\left(4x+1\right)\)
41: Ta có: \(A=8x^2+10x-3\)
\(=8x^2+12x-2x-3\)
\(=4x\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(2x+3\right)\left(4x-1\right)\)
42: Ta có: \(A=-8x^2+23x+3\)
\(=-8x^2+24x-x+3\)
\(=-8x\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(-8x-1\right)\)
43: Ta có: \(A=8x^2-23x-3\)
\(=8x^2-24x+x-3\)
\(=8x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(8x+1\right)\)
44: Ta có: \(A=10x^2-11x-6\)
\(=10x^2-15x+4x-6\)
\(=5x\left(2x-3\right)+2\left(2x-3\right)\)
\(=\left(2x-3\right)\left(5x+2\right)\)
45: Ta có: \(A=-10x^2+11x+6\)
\(=-10x^2+15x-4x+6\)
\(=-5x\left(2x-3\right)-2\left(2x-3\right)\)
\(=\left(2x-3\right)\left(-5x-2\right)\)
\(A=x^7-4x^3+x^2+2=x^3\left(x^4-4\right)+x^2+2\)
\(=x^3\left(x^2-2\right)\left(x^2+2\right)+x^2+2\)
\(=\left(x^2+2\right)\left(x^3\left(x^2-2\right)+1\right)\)
\(=\left(x^2+2\right)\left(x^5-2x^3+1\right)\)
\(=\left(x^2+2\right)\left(x^5-x^4+x^4-x^3-x^3+x^2-x^2+x-x+1\right)\)
\(=\left(x^2+2\right)\left[x^4\left(x-1\right)+x^3\left(x-1\right)-x^2\left(x-1\right)-x\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(x^2+2\right)\left(x-1\right)\left(x^4+x^3-x^2-x-1\right)\)