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a)\(A=x^2-1\)
\(Nx:\)\(x^2\ge0\)
\(\Rightarrow A_{Min}=0-1=-1\Leftrightarrow x=0\)
b) \(B=x^2-2x+3\)
\(=x\left(x-2\right)+3\)
\(Nx:x\left(x-2\right)\ge0\)
\(\Rightarrow B_{Min}=3\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow x=0\)
c) \(C=\left|2x+1\right|-5\)
\(Nx:\left|2x+1\right|\ge0\Rightarrow2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)
\(\Rightarrow C_{Min}=-5\Leftrightarrow x=\frac{-1}{2}\)
d) \(D=3x^2+6x-7\)
\(=3\left(x^2+2x\right)-7\)
\(Nx:Min_{x^2+2x}=-1\Leftrightarrow x=-1\)
\(D_{Min}=-8\Leftrightarrow x=-1\)
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x^2-25x^4=0
=>x^2-25x^2.x^2=0
=>x^2.(1-25x^2)=0
=>x=0 hoặc x^2=1/25
=>x thuộc {-0,2;0;0,2}
2) 2 giá trị
3)x^2+7x+12=0
=>x^2+3x+4x+3.4=0
=>x(x+3)+4(x+3)=0
=>(x+4)(x+3)=0
=>x=-3;x=-4
nhớ ****
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1)x thuộc {-0,2;0;0,2}
2)2 giá trị
3)x^2+3x+4x+4.3=0
=>x(x+3)+4(x+3)=0
=>(x+3)(x+4)=0
=>x=-4;x=-3
1)x2-25x4=0
x2(1-25x2)=0
=>x^2=0 hoặc 1-25x^2=0
x=0 25x^2=-1-0=1
x^2=1/25=(1/5)^2=(1/-5)^2
Vậy S={-1/5;0;1/5}
2)Có 3 giá trị là 0;1;2
3)có 2 giá trị là -3;-4
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d)
\(\left(\frac{7^3\left(7-1\right)}{7^6}\right)^2\)
\(=\left(\frac{6}{7^3}\right)^2\)
\(=\frac{6^2}{7^{3^2}}\)
\(=\frac{36}{7^6}\)
\(\left(\frac{7^3\left(7-1\right)}{7^6}\right)^2\)
\(=\left(\frac{6}{7^3}\right)^2\)
\(=\left(\frac{6^2}{7^{3^2}}\right)\)
\(=\frac{36}{7^6}\)
Code : Breacker
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1.\(x\left(x+y\right)=-45;y\left(x+y\right)=5\Rightarrow\left(x+y\right)\left(x+y\right)=-45+5=-40\Rightarrow\left(x+y\right)^2=-40\Rightarrow\left(x+y\right)\varepsilon\phi\Rightarrow x,y\varepsilon\phi\)