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\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\cdot\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}+1}\right)=1-a\) ( với đk a ≥ 0 và a ≠ 1)
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\(\left(3x-3\right)^2+\left(4y+2\right)^2=0\)
Ta có:
`(3x-3)^2>=0` với mọi x
`(4y+2)^2>=0` với mọi x
`=>(3x-3)^2+(4y+2)^2>=0` với mọi x,y
Mặt khác: `(3x-3)^2+(4y+2)^2=0`
Dấu "=" xảy ra: `3x-3=0` và `4y+2=0`
`=>3x=3` và `4y=-2`
`=>x=3/3=1` và `y=-2/4=-1/2`
3 x 9 + 18 x 2 + 2 x 9 + 9
= 3 x 9 + 9 x 2 x 2 + 2 x 9 + 9
= 9 x (3 + 2 + 2 + 2 + 1)
= 9 x 10
= 90
\(=3\times9+9\times2\times2+2\times9+9\)
\(=3\times9+4\times9+2\times9+9\)
\(=9\times\left(3+4+2+1\right)\)
\(=9\times10\)
\(=90\)
C1:
\(A=\left\{0;1;2;3;4;5;6;7;8;9;10;11;12\right\}\)
\(C2:A=\left\{x\inℕ|x\le12\right\}\)
a.
\(\sqrt{x^2-4x+1}=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2-4x+1=x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-4x+1=0\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{1}{4}\)
b.
\(\sqrt{5x^2-2x+2}=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\5x^2-2x+2=\left(x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\4x^2-4x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{1}{2}\)
c.
\(\sqrt{x^2-8x+16}=4-x\)
\(\Leftrightarrow\sqrt{\left(4-x\right)^2}=4-x\)
\(\Leftrightarrow\left|4-x\right|=4-x\)
\(\Leftrightarrow4-x\ge0\)
\(\Rightarrow x\le4\)
d.
\(\sqrt{3x+1}=\sqrt{4x-3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x-3\ge0\\3x+1=4x-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\end{matrix}\right.\)
\(\Rightarrow x=4\)
\(15)15+4\left(x-2\right)=95\\ 4\left(x-2\right)=95-15\\ 4\left(x-2\right)=80\\ x-2=\dfrac{80}{4}\\ x-2=20\\ x=20+2\\ x=22\\ 16)20-\left(x+14\right)=5\\ x+14=20-5\\ x+14=15\\ x=15-14\\ x=1\\ 17)24+3\left(5-x\right)=27\\ 3\left(5-x\right)=27-24\\ 3\left(5-x\right)=3\\ 5-x=\dfrac{3}{3}=1\\ x=5-1=4\\ 18)15:x=5\\ x=15:5\\ x=3\\ 19)\dfrac{x}{4}=3\\ x=4\cdot3\\ x=12\\ 20)\dfrac{21}{x}=7\\ x=\dfrac{21}{7}\\ x=3\)
\(A=10\cdot\dfrac{4}{7}-1,38-7,62\\ =10\cdot\dfrac{4}{7}-\left(1,38+7,62\right)\\ =\dfrac{10\cdot4}{7}-9\\ =\dfrac{40}{7}-\dfrac{63}{7}\\ =\dfrac{40-63}{7}\\ =\dfrac{-23}{7}\)
\(1)A=x^2-7x+2\\ =\left(x^2-2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}\right)-\dfrac{41}{4}\\ =\left(x-\dfrac{7}{2}\right)^2-\dfrac{41}{4}\)
Ta có: `(x-7/2)^2>=0` với mọi x
`=>A=(x-7/2)^2-41/4>=-41/4` với mọi x
Dấu "=" xảy ra: `x-7/2=0<=>x=7/2`
\(2)B=9x^2-12x+5\\ =\left(9x^2-12x+4\right)+1\\ =\left[\left(3x\right)^2-2\cdot3x\cdot2+2^2\right]+1\\ =\left(3x-2\right)^2+1\)
Ta có: `(3x-2)^2>=0` với mọi x
`=>B=(3x-2)^2+1>=1` với mọi x
Dấu "=" xảy ra: `3x-2=0<=>x=2/3`
`23*16+23*84-300`
`=23*(16+84)-300`
`=23*100-300`
`=2300-300`
`=2000 `
\(VT=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\cdot\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\\ =\left[1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right]\cdot\left[1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right]\\ =\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\\ =1-\left(\sqrt{a}\right)^2\\ =1-a=VP\)