5x² - 4(x² - 2x + 1) - 5 = 0

=> 5x² - 4x² + 8x - 4 - 5 = 0 

=> x² + 8x - 9 = 0

=> x² + 9x - x - 9 = 0 

=> x(x + 9) - (x + 9) = 0

=> (x + 9)(x - 1) = 0

=> x + 9 = 0 => x = -9

hoặc x - 1 = 0 = > x = 1

                                                                       Vậy x = -9, x = 1

\(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\left(5x^2-5\right)-4\left(x^2-2.1.x+1^2\right)=0\)

\(5\left(x^2-1\right)-4\left(x-1\right)^2=0\)

\(5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)\left(x-1\right)=0\)

\(\left[5\left(x+1\right)-4\left(x-1\right)\right]\left(x-1\right)=0\)

\(\left(5x+5-4x+4\right)\left(x-1\right)=0\)

\(\left(x+9\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+9=0\\x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-9\\x=1\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=-9\\x=1\end{cases}}.\)

1-x-2x^2

= 1-x-2x.2x

= 1 - ( x + 2x.2x)

= 1 - 5x

Để 1-x-2x^2 mang giá trị lớn nhất thì x phài là số âm.

\(A=1-x-2x^2\)

\(=-2\left(x^2+2\times x\times\frac{1}{4}+\left(\frac{1}{4}\right)^2-\left(\frac{1}{4}\right)^2-\frac{1}{2}\right)\)

\(=-2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]\)

\(\left(x+\frac{1}{4}\right)^2\ge0\)

\(\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\ge-\frac{9}{16}\)

\(-2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]\le\frac{9}{8}\)

Vậy Max A = \(\frac{9}{8}\) khi x = \(-\frac{1}{4}\)

\(A=-2x^2-x+1=-\left(2x^2+\sqrt{2}.2.\frac{1}{2\sqrt{2}}x+\frac{1}{8}\right)+\frac{9}{8}\)

\(A=-\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2+\frac{9}{8}\le\frac{9}{8}\)

\(MaxA=\frac{9}{8}\Leftrightarrow\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2=0\Leftrightarrow x=-\frac{1}{4}\)

Vì đa thức \(f\left(x\right)=2x^2+\text{ax}+b-2.\)chia hết cho \(x-1;x+2\)

Theo đinh lí Bơ - ru ta có 

\(f\left(1\right)=2.1^2+a+b-2=0\Rightarrow a+b=0\) (1)

\(f\left(-2\right)=2.\left(-2\right)^2-2a+b-2=0\Rightarrow6-2a+b=0\Rightarrow2a-b=6\)(2)

Cộng (1) và (2) suy ra 

\(3a=6\Rightarrow a=2\)thay vào a+b=0 ta có : \(2+b=0\Rightarrow b=-2\)

Vậy a=2 ; a=-2

a)x^m+4+x^m+3-x-1

=(x^m+4-x)+(x^m+3-1)

=x(x^m+3-1)+(x^m+3-1)

=(x+1)(x^m+3-1)

Với x=-2 ta có:... bn tự thay

b)x⁶-x⁴+2x³+2x²=x⁶-2x⁵+2x⁴+2x⁵-4x⁴+4x³+x⁴-2x³+2x²

=x⁴(x²-2x+2)+2x³(x²-2x+2)+x²(x²-2x+2)

=(x⁴+2x³+x²)(x²-2x+2)

=[x²(x²+2x+1)](x²-2x+2)

=x²(x+1)²(x²-2x+2)

Với x=-2 bn tự thay nhé h mk bận

\(2x^2-x-15\)

\(=\left(x-3\right)\left(x+\frac{5}{2}\right)\)

\(x^4+x^2+1\)

a)   \(2x^2-x-15=2x^2-6x+5x-15=2x\left(x-3\right)+5\left(x-3\right)=\left(x-3\right)\left(2x-5\right)\)

a)\(2x^2-x-15=2x^2-6x+5x-15\)

\(=2x\left(x-3\right)+5\left(x-3\right)\)

\(=\left(x-3\right)\left(2x+5\right)\)

\(x^4+x^2+1=x^4+x^3+x^2-x^3+1\)

\(=x^2\left(x^2+x+1\right)-\left(x^3-1\right)\)

\(=x^2\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)

\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)