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Đây nhé bn
B(3)=2*3^2-4*3+3=18-12+3=9
B(-1/2)=2*1/4-4*(-1/2)+3=1/2+3+2=1/2+5=11/2
a) \(({x^2} + 2x + 3) + (3{x^2} - 5x + 1) = ({x^2} + 3{x^2}) + (2x - 5x) + (3 + 1) = 4{x^2} - 3x + 4\);
b) \(\begin{array}{l}(4{x^3} - 2{x^2} - 6) - ({x^3} - 7{x^2} + x - 5) = 4{x^3} - 2{x^2} - 6 - {x^3} + 7{x^2} - x + 5\\ = (4{x^3} - {x^3}) + ( - 2{x^2} + 7{x^2}) - x + ( - 6 + 5) = 3{x^3} + 5{x^2} - x - 1\end{array}\);
c) \(\begin{array}{l} - 3{x^2}(6{x^2} - 8x + 1) = - 3{x^2}.6{x^2} - - 3{x^2}.8x + - 3{x^2}.1\\ = - 18{x^{2 + 2}} + 24{x^{2 + 1}} - 3{x^2} = - 18{x^4} + 24{x^3} - 3{x^2}\end{array}\);
d) \(\begin{array}{l}(4{x^2} + 2x + 1)(2x - 1) = (4{x^2} + 2x + 1).2x - (4{x^2} + 2x + 1).1 = 4{x^2}.2x + 2x.2x + 1.2x - 4{x^2} - 2x - 1\\ = 8{x^{2 + 1}} + 4{x^{1 + 1}} + 2x - 4{x^2} - 2x - 1 = 8{x^3} + 4{x^2} + 2x - 4{x^2} - 2x - 1 = 8{x^3} - 1\end{array}\);
e) \(\begin{array}{l}({x^6} - 2{x^4} + {x^2}):( - 2{x^2}) = {x^6}:( - 2{x^2}) - 2{x^4}:( - 2{x^2}) + {x^2}:( - 2{x^2})\\ = - \dfrac{1}{2}{x^{6 - 2}} + {x^{4 - 2}} - \dfrac{1}{2}{x^{2 - 2}} = - \dfrac{1}{2}{x^4} + {x^2} - \dfrac{1}{2}.\end{array}\);
g)
\(({x^5} - {x^4} - 2{x^3}):({x^2} + x)=x^3-2x^2\)
a: =>(3/2-2x):2/3=1/6
=>3/2-2x=1/6x2/3=2/18=1/9
=>2x=25/18
hay x=25/36
b: \(\Leftrightarrow2x-2x+\dfrac{5}{2}-2=x-\dfrac{1}{4}\)
=>x-1/4=1/2
=>x=3/4
c: \(\Leftrightarrow2x-\dfrac{2}{3}-\dfrac{1}{3}x+\dfrac{1}{4}x=0\)
=>23/12x=2/3
=>x=8/23
1) 
\(\left|x+\frac{4}{5}\right|+\frac{7}{5}=\frac{3}{5}\)
\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{3}{5}-\frac{7}{5}\)
\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{-4}{5}\)
\(x+\frac{4}{5}=\pm\frac{4}{5}\)
\(TH1:x+\frac{4}{5}=\frac{4}{5}\)
\(\Rightarrow x=\frac{4}{5}-\frac{4}{5}=0\)
\(TH2:x+\frac{4}{5}=\frac{-4}{5}\)
\(\Rightarrow x=\frac{-4}{5}-\frac{4}{5}=\frac{-8}{5}\)
Vậy x ∈ {0; \(\frac{-8}{5}\)}
1) \(\left(x+2\right)^3-\left(x+6\right)^2-\left(x+1\right)\left(x^2-x+1\right)\)
\(=x^3+3.x^2.2+3.x.2^2+2^3-\left(x^2+2.x.6+6^2\right)-\left(x^3+1\right)\)
\(=x^3+6x^2+12x+8-x^2-12x-36-x^3-1\)
\(=5x^2-29\)
2. \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x+3\right)^2\)
\(=x^3-3^3-\left(x^2+2.x.3+3^2\right)\)
\(=x^3-27-x^2-6x-9\)
\(=x^3-x^2-6x-36\)
\(=x^3-2x^2+3x^2-6x-36\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)-36\)
\(=x\left(x-2\right)\left(x+3\right)-36\)
3. \(\left(2x-1\right)^2+2x^2.\left(x-2\right)\left(x+3\right)^2\)
\(=4x^2-4x+1+2x^2\left(x-2\right)\left(x^2+6x+9\right)\)
\(=4x^2-4x+1+2x^2\left(x^3+6x^2+9x-2x^2-12x-18\right)\)
\(=4x^2-4x+1+2x^2\left(x^3+4x^2-3x-18\right)\)
\(=4x^2-4x+1+2x^5+8x^4-6x^3-36x^2\)
\(=2x^5+8x^4-6x^3-32x^2-4x+1\)
....
P/s: Không chắc lắm
Hai câu cuối khó nhìn nên ko giải