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cho S=1-3+32-33+...+398-399
a. Chứng minh: S chia hêt cho 20
b. Rút gọn S, từ đó suy ra 3100 chia 4 dư 1
chịu
Dat \(a=\sqrt[3]{65+x},b=\sqrt[3]{65-x}\)
Bien doi PT thanh \(a^2+4b^2=5ab\)
\(\Leftrightarrow a^2-5ab+4b^2=0\)
\(\Leftrightarrow\left(a^2-ab\right)-\left(4ab-4b^2\right)=0\)
\(\Leftrightarrow a\left(a-b\right)-4b\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\left(1\right)\\a=4b\left(2\right)\end{cases}}\)
\(\left(1\right)\Leftrightarrow\sqrt[3]{65+x}=\sqrt[3]{65-x}\)
\(\Leftrightarrow65+x=65-x\)
\(\Leftrightarrow x=0\left(n\right)\)
\(\left(2\right)\Leftrightarrow\sqrt[3]{65+x}=4\sqrt[3]{65-x}\)
\(\Leftrightarrow65+x=64.65-64x\)
\(\Leftrightarrow65x=64.65-65\)
\(\Leftrightarrow x=63\left(n\right)\)
Vay nghiem cua PT la \(x=0,x=63\)
\(D=\frac{2}{\sqrt{xy}}:\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}\right)^2-\frac{x+y}{x-2\sqrt{xy}+y}\left(ĐKXĐ:x\ge0,y\ge0,x\ne y\right)\)
\(\Leftrightarrow D=\frac{2}{\sqrt{xy}}:\left(\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}\right)^2-\frac{x+y}{\sqrt{x}}\)
\(\Leftrightarrow D=\frac{2}{\sqrt{xy}}.\frac{xy}{\left(\sqrt{x}-\sqrt{y}\right)^2}-\frac{x+y}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
\(\Leftrightarrow D=\frac{2\sqrt{xy}-x-y}{\left(\sqrt{x}-\sqrt{y}\right)^2}=\frac{-\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}-\sqrt{y}\right)^2}=-1\)
=> ko phụ thuộc x
a)\(-\frac{2}{\sqrt{1-3x}}\text{có nghĩa }\Leftrightarrow1-3x>0\)
\(\Leftrightarrow-3x>-1\Leftrightarrow x< 1\)
b)\(\sqrt{\frac{-5}{x^2+6}}\text{có nghĩa }\Leftrightarrow\frac{-5}{x^2+6}\ge0;x^2+6\ne0\)
\(\Leftrightarrow x^2+6< 0\Leftrightarrow x^2< -6\left(\text{vô lí }\right)\)
\(x\in\varnothing\)
\(\sqrt{x+5}+\frac{1}{x+5}\text{có nghĩa }\Leftrightarrow x+5>0\)
\(\Leftrightarrow x>-5\)
\(\sqrt{\left(x-1\right)\left(x-2\right)}\text{có nghĩa }\Leftrightarrow\left(x-1\right)\left(x-2\right)\ge0\)
TH1: \(\left(x-1\right)\ge0\text{ và }\left(x-2\right)\ge0\)
\(\Rightarrow x\ge2\)
TH2: \(\left(x-1\right)\le0\text{ và }\left(x-2\right)\le0\)
\(\Rightarrow x\le1\)
\(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)
\(\Leftrightarrow x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019^2\)
\(\Leftrightarrow M=2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019^2-2x^2y^2-x^2-y^2-1\)(Đặt M = .... cho gọn)
Có \(S=x\sqrt{1+y^2}+y\sqrt{1+x^2}\)
\(\Rightarrow S^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+M\)
\(\Rightarrow S^2=2x^2y^2+x^2+y^2+2019^2-2x^2y^2-x^2-y^2-1\)
\(\Rightarrow S=\sqrt{2019^2-1}\)