Question

a) tìm x,y,z biết:\(\left(2x-1\right)^{ }+\left(y-\dfrac{2}{5}\right)^{ }+|x+y-z|=0\)

b)so sánh \(\sqrt{15}+\sqrt{35}\) và\(2\sqrt{26}\)

c)so sánh 2¹²⁶và 3⁸⁴

Answer 1

b)Ta có:\(\sqrt{15}< \sqrt{16}=4\)

\(\sqrt{35}< \sqrt{36}=6\)

\(\Rightarrow\sqrt{15}+\sqrt{35}< 10\)(1)

Mà \(\sqrt{26}>\sqrt{25}=5\) \(\Rightarrow2\sqrt{26}>10\left(2\right)\)

Từ (1),(2)\(\Rightarrow\sqrt{15}+\sqrt{35}< 2\sqrt{26}\)

Answer 2

c)Ta có:\(2^{126}=\left(2^3\right)^{42}=8^{42}\)

\(3^{84}=\left(3^2\right)^{42}=9^{42}\)

Vì \(8^{42}< 9^{42}\) nên \(2^{126}< 3^{84}\)

Answer 3

Bùi Hà Chi giúp mình câu a) đi

Answer 4

b) \(\sqrt{15}+\sqrt{35}< \sqrt{16}+\sqrt{36}=4+6=10=2.5=2\sqrt{25}< 2\sqrt{26}\)

Vậy \(\sqrt{15}+\sqrt{35}< 2\sqrt{26}\)

Answer 5

a) để \(\)\(\left(2x-1\right)+\left(y-\dfrac{2}{5}\right)+\left|x+y-z\right|\)

\(\Rightarrow\left\{{}\begin{matrix}x+y-z=0\\y-\dfrac{2}{5}=0\\2x-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y-z=0\\y=\dfrac{2}{5}\\x=\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}+\dfrac{2}{5}=z\\y=\dfrac{2}{5}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}z=\dfrac{9}{10}\\y=\dfrac{2}{5}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy ...