\(B=y^2-y+1\)

\(=y^2-2.y.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Vì \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall y\Rightarrow B\ge\dfrac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow y=\dfrac{1}{2}\)

\(C=x^2-4x+y^2-y+5\)

\(=x^2-4x+4+y^2-y+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2\)

Vì \(\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2\ge0\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{100}\)

\(=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(2x=5.9^2:8\)

\(2x=5.81:8\)

\(16x=405\)

\(x=\dfrac{405}{16}\)

Mình vào từ 18/3/2021, lúc đó mình tra bài trên mạng và tình cờ biết đến web.

\(c=\sqrt{a^2+b^2-2.a.b.cosC}\)

\(=\sqrt{49,4^2+26,4^2-2.26,4.49,4.cos47^o20'}\simeq37\)

Ta có:

\(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{\left(26,4\right)^2+37^2-\left(49,4\right)^2}{2.26,4.37}\simeq-0,2\)

\(\Rightarrow\widehat{A}\simeq101,5^o\)

\(\Rightarrow\widehat{B}=180^o-101,5^o-47,3^o=31,2^o\)

\(x^2+2y^2-4x+2y+\dfrac{9}{2}=0\)

\(x^2-4x+4+2y^2+2y+\dfrac{1}{2}=0\)

\(\left(x-2\right)^2+2\left(y+\dfrac{1}{2}\right)^2=0\)

Vì \(\left(x-2\right)^2+2\left(y+\dfrac{1}{2}\right)^2\ge0\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}\end{matrix}\right.\)

a) \(4x+4=16\)

\(4x=12\)

\(x=3\)

b) \(34\left(2x-6\right)=0\)

\(2x=6\)

\(x=3\)

c) \(15:x=5\)

\(x=15:5=3\)

d) \(20-\left(x+14\right)=5\)

\(x+14=20-5=15\)

\(x=15-14=1\)

a) \(n\inƯ\left(20\right)=\left\{1;2;4;5;10;20\right\}\)

b) \(2n+1\inƯ\left(18\right)=\left\{1;2;3;6;9;18\right\}\)

=> \(n\in\left\{0;1;4\right\}\)

c) \(n\left(n+2\right)=8\)

\(\left(n+1\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}n+1=3\\n+1=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}n=2\left(TM\right)\\m=-4\left(L\right)\end{matrix}\right.\)

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