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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
![](https://rs.olm.vn/images/avt/0.png?1311)
a: A(x)=0
=>9x=-13
=>x=-13/9
b: x^2-49=0
=>x^2=49
=>x=7 hoặc x=-7
c: x^2-7=0
=>x^2=7
=>x=căn 7 hoặc x=-căn 7
d: 2x^2-32=0
=>x^2-16=0
=>x^2=16
=>x=4 hoặc x=-4
e: 3x^2-5=0
=>3x^2=5
=>x^2=5/3
=>\(x=\pm\sqrt{\dfrac{5}{3}}\)
g: x^2+6x=0
=>x(x+6)=0
=>x=0 hoặc x=-6
m: M(x)=0
=>5x(x-2)=0
=>x=0 hoặc x=2
n: x^3-9x=0
=>x(x^2-9)=0
=>x(x-3)(x+3)=0
=>x=0;x=3;x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
\(({x^5} - {x^4} - 2{x^3}):({x^2} + x)=x^3-2x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{x}{x+1}=\frac{1}{2}\)
=> 2x = x + 1
=> 2x - x = 1
=> x = 1
b) \(\frac{x}{2}=\frac{x}{3}\)
=> 3x = 2x
=> 3x - 2x = 0
=> x = 0
c) \(\frac{x+1}{2}=\frac{x+1}{2017}\)
=> \(2017\left(x+1\right)=2\left(x+1\right)\)
=> 2017x + 2017 = 2x + 2
=> 2017x - 2x = 2 - 2017
=> 2015x = -2015
=> x = -2015 : 2015
=> x = -1
i) \(\frac{3}{x}=\frac{x}{2017}\)
=> x2 = 2017.3
=> x2 = 6051
=> \(\orbr{\begin{cases}x=\sqrt{6051}\\x=-\sqrt{6051}\end{cases}}\)
còn lại tự lm
\(a,\frac{x}{x+1}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow x=\frac{1}{2}x+\frac{1}{2}\)
\(\Rightarrow x-\frac{1}{2}x=\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}x=\frac{1}{2}\)
\(\Rightarrow x=1\)
\(b,\frac{x}{2}=\frac{x}{3}\)
\(\Rightarrow x=\frac{x}{3}.2\)
\(\Rightarrow x=\frac{2x}{3}\)
\(\Rightarrow3x=2x\)
\(\Rightarrow x=0\)
\(c,\frac{x+1}{2}=\frac{x+1}{2017}\)
\(\Rightarrow x+1=\frac{x+1}{2017}.2\)
\(\Rightarrow x+1=\frac{2x+2}{2017}\)
\(\Rightarrow2017x+2017=2x+2\)
\(\Rightarrow2017x-2x=2-2017\)
\(\Rightarrow2015x=-2015\)
\(\Rightarrow x=-1\)
\(i,\frac{3}{x}=\frac{x}{2017}\)
\(\Rightarrow x=3:\frac{x}{2017}\)
\(\Rightarrow x=\frac{6051}{x}\)
\(\Rightarrow x^2=6051\)
\(\Rightarrow x=\sqrt{6051}\)
\(o,\frac{x}{3}=\frac{x+1}{2}\)
\(\Rightarrow x=\frac{x+1}{2}.3\)
\(\Rightarrow x=\frac{3x+3}{2}\)
\(\Rightarrow2x=3x+3\)
\(\Rightarrow-x=3\)
\(\Rightarrow x=-3\)
\(m,\frac{x+1}{2}=\frac{x+2}{3}\)
\(\Rightarrow x+1=\frac{x+2}{3}.2\)
\(\Rightarrow x+1=\frac{2x+4}{3}\)
\(\Rightarrow3x+3=2x+4\)
\(\Rightarrow x=1\)
\(p,\frac{x+1}{2}=x\)
\(\Rightarrow2x=x+1\)
\(\Rightarrow x=1\)
\(m,\frac{2}{x}=\frac{x}{8}\)
\(\Rightarrow x=2:\frac{x}{8}\)
\(\Rightarrow x=\frac{16}{x}\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=4\)
\(Q,\frac{x^2}{2}=\frac{8}{x^2}\)
\(\Rightarrow x^2=\frac{8}{x^2}.2\)
\(\Rightarrow x^2=\frac{16}{x^2}\)
\(\Rightarrow x^4=16\)
\(\Rightarrow x=2\)
\(r,\frac{x^3}{2}=\frac{32}{x}\)
\(\Rightarrow x^3=\frac{32}{x}.2\)
\(\Rightarrow x^3=\frac{64}{x}\)
\(\Rightarrow x^4=64\)
\(\Rightarrow x=\sqrt[4]{64}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)
=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0
=>6x-24=0
=>x=4
e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2
=>-5x^2-2x+16+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
\(x^2=x^3\)
=> \(x^3-x^2=0\)
=> \(x^2\left(x-1\right)=0\)
=> \(\orbr{\begin{cases}x^2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Tính x hộ mik nha