Bài 1: So sánh
A= (2+1)*(2^2+1)*(2^4+1)*(2^16+1)
B= (2^23-1)
Bài 2: Cho x-y=5 và y*x=14
Tính A= x^2+y^2
B= (x+y)^2
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3:
a: \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)
b: \(\dfrac{x}{y}\cdot\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}\cdot\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)
2:
a: 2căn 7=căn 28
3căn 2=căn 18
mà 28>18
nên 2*căn 7>3*căn 2
b: 5=2+3
mà 3>căn 2
nên 2+3>2+căn 2
=>5>2+căn 2
1) a) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)
\(=\sqrt{49.2}-\sqrt{36.2}+0,5\sqrt{4.2}\)
\(=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}\)
\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
b) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49}\)
\(=3\sqrt{a}-4\sqrt{a}+7=7-\sqrt{a}\)
2. a) \(2\sqrt{7}=\sqrt{4.7}=\sqrt{28}\)
\(3\sqrt{2}=\sqrt{9.2}=\sqrt{18}\)
Mà \(\sqrt{28}>\sqrt{18}\Rightarrow2\sqrt{7}>3\sqrt{2}\)
b) \(5=2+3=2+\sqrt{9}\)
Vì \(\sqrt{9}>\sqrt{2}\Rightarrow2+\sqrt{9}>2+\sqrt{2}\Rightarrow5>2+\sqrt{2}\)
3. a) \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)
b) \(\dfrac{x}{y}.\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}.\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)
bài 1:
a) (x+1)^2-(x-1)^2-3(x+1)(x-1)
=(x+1+x-1)(x+1-x+1)-3x^2-3
=2x^2-3x^2-3
=-x^2-3
Bài 1 :
\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)
\(\Leftrightarrow x=-23\)
\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)
1/
\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^{16}+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^{16}+1\right)=\left(2^4-1\right)\left(2^4+1\right)\left(2^{16}-1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)
Vì 232 > 223 => 232-1>223-1 hay A>B
2/
\(A=x^2+y^2=\left(x-y\right)^2+2xy=5^2+2.14=25+28=53\)
\(B=\left(x+y\right)^2=\left(x-y\right)^2+4xy=5^2+4.14=25+56=81\)