Question

Giúp mình bài 1,3,4 ạ

Answer 1

Bài 3:

a: \(\sqrt{10+43+91}=\sqrt{101+43}=\sqrt{144}=12\)

b: \(\sqrt{\dfrac{1}{9}}\cdot\sqrt{0,81}+2^3\cdot\sqrt{0,09}\)

\(=\dfrac{1}{3}\cdot0,9+8\cdot0,3\)

=0,3+2,4

=2,7

c: \(\left(\dfrac{2}{5}\sqrt{16}+2\cdot\sqrt{\dfrac{16}{25}}\right):2\sqrt{\dfrac{1}{16}}\)

\(=\left(\dfrac{2}{5}\cdot4+2\cdot\dfrac{4}{5}\right):\left(2\cdot\dfrac{1}{4}\right)\)

\(=\left(\dfrac{8}{5}+\dfrac{8}{5}\right):\dfrac{1}{2}=\dfrac{16}{5}\cdot2=\dfrac{32}{5}\)

d: \(\left(\sqrt{\dfrac{16}{25}}-\sqrt{\dfrac{9}{49}}+\sqrt{\dfrac{36}{25}}\right):\sqrt{\dfrac{121}{196}}\)

\(=\left(\dfrac{4}{5}-\dfrac{3}{7}+\dfrac{6}{5}\right):\dfrac{11}{14}=\left(2-\dfrac{3}{7}\right)\cdot\dfrac{14}{11}\)

\(=\dfrac{11}{7}\cdot\dfrac{14}{11}=\dfrac{14}{7}=2\)

Bài 4:

a: ĐKXĐ: x>=0

\(2\sqrt{x}-1=17\)

=>\(2\sqrt{x}=18\)

=>\(\sqrt{x}=\dfrac{18}{2}=9\)

=>\(x=9^2=81\)

b: ĐKXĐ: x>=0

\(\left(3\sqrt{x}-1\right)^2=\dfrac{64}{25}\)

=>\(\left[{}\begin{matrix}3\sqrt{x}-1=\dfrac{8}{5}\\3\sqrt{x}-1=-\dfrac{8}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x}=\dfrac{13}{5}\\3\sqrt{x}=-\dfrac{3}{5}\left(loại\right)\end{matrix}\right.\)

=>\(\sqrt{x}=\dfrac{13}{5}:3=\dfrac{13}{15}\)

=>\(x=\left(\dfrac{13}{15}\right)^2=\dfrac{169}{225}\left(nhận\right)\)

c: ĐKXĐ: x>=0

\(\left(x^2+5\right)\left(x-6\sqrt{x}\right)=0\)

mà \(x^2+5>=5>0\forall x\)

nên \(x-6\sqrt{x}=0\)

=>\(\sqrt{x}\left(\sqrt{x}-6\right)=0\)

=>\(\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=36\left(nhận\right)\end{matrix}\right.\)