\(\sqrt{x^2+2}+\sqrt{x^2+2x+3}=1\)
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a/ \(x^2+y^2=x^2+y^2+2xy-2xy\)\(=\left(x+y\right)^2-2xy\)
thay vào: \(\left(x+y\right)^2-2xy=a^2-2b\)
b/ \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=\left(x+y\right)\left(x^2+y^2+2xy-xy-2xy\right)\)\(=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)
thay vào: \(=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]=a\left(a^2-3b\right)\)
c/ \(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=\left[\left(x+y\right)^2-2xy\right]^2-2x^2y^2\)
thay vào: \(\left[\left(x+y\right)^2-2xy\right]^2-2x^2y^2=\left(a^2-2b\right)^2-2b^2\)
a) \(-x^2+6x+1=-\left(x^2-6x+9\right)+10=-\left(x-3\right)^2+10\le10\)
Vậy Max = 10 <=> x = 3
b) \(-5x^2-4x+1=-5\left(x^2+2.x.\frac{2}{5}+\frac{4}{25}\right)+\frac{4}{5}+1=-5\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\le\frac{9}{5}\)
Vậy Max = \(\frac{9}{5}\Leftrightarrow x=-\frac{2}{5}\)
a/ -4x(x - 5) - 2x(8 - 2x) = -3
=> -4x2 + 20x - 16x + 4x2 = -3
=> 4x = -3
=> x = -3/4
b/ \(\frac{x-1}{-15}=-\frac{60}{x-1}\Rightarrow\left(x-1\right)^2=\left(-60\right)\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900\Rightarrow\orbr{\begin{cases}x-1=30\\x-1=-30\end{cases}\Rightarrow\orbr{\begin{cases}x=31\\x=-29\end{cases}}}\)
Vậy x = -29 , x = 31
a/ \(2x^2+8x+1=2\left(x^2+4x+\frac{1}{2}\right)=2\left(x^2+2.2x+4-4+\frac{1}{2}\right)\)
\(=2\left[\left(x+2\right)^2-\frac{7}{2}\right]=2\left(x+2\right)^2-7\ge-7\)
Vậy Min A = -7 khi x + 2 = 0 => x = 2
b/ \(2x^2+3x+1=2\left(x^2+\frac{3}{2}x+\frac{1}{2}\right)=2\left(x^2+2.\frac{3}{4}.x+\frac{9}{16}-\frac{9}{16}+\frac{1}{2}\right)\)
\(=2\left[\left(x+\frac{3}{4}\right)^2-\frac{1}{16}\right]=2\left(x+\frac{3}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)
Vậy Min B = -1/8 khi x + 3/4 = 0 => x = -3/4
2x + 2x+1 + 2x+2 + 2x+3 = 120
=> 2x ( 1 + 2 + 22 + 23) = 120
=> 2x . 15 = 120
=> 2x = 8 = 23
=> x = 3
\(2^x+2^x\cdot2^1+2^x\cdot2^2+2^x\cdot2^3=120\)
\(2^x\cdot\left(1+2+4+8\right)=120\)
\(2^x\cdot15=120\)
\(2^x=120:15\)
\(2^x=8\)
\(2^x=2^3\)
\(=>x=3\)
\(x^2+\frac{1}{x^2}=7\Leftrightarrow x^2+2+\frac{1}{x^2}=9\Leftrightarrow\left(x+\frac{1}{x}\right)^2=3^2.\)Do x > 0 nên \(x+\frac{1}{x}\)>0 và \(x+\frac{1}{x}=3\)
\(\Rightarrow\left(x+\frac{1}{x}\right)^3=27\Rightarrow x^3+\frac{1}{x^3}+3\cdot x\cdot\frac{1}{x}\left(x+\frac{1}{x}\right)=27\Rightarrow x^3+\frac{1}{x^3}+3\cdot3=27\Rightarrow x^3+\frac{1}{x^3}=18\)
\(\Rightarrow\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=7\cdot18\Rightarrow x^5+\frac{1}{x^5}+x+\frac{1}{x}=126\Rightarrow x^5+\frac{1}{x^5}+3=126\Rightarrow x^5+\frac{1}{x^5}=123.\)
Vậy \(x^5+\frac{1}{x^5}\)là 1 số nguyên và bằng: 123
\(\sqrt{x^2+2}\ge\sqrt{2}>1\forall x.\)
\(\Rightarrow\sqrt{x^2+2}+\sqrt{x^2+2x+3}>1\forall x\)
Dấu "=" không xảy ra nên PT vô nghiệm.