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\(\hept{\begin{cases}\left(\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25}\right)^2=9\left(\sqrt[3]{5}-\sqrt[3]{4}\right)\\\left(3\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}\right)^2=9\left(\sqrt[3]{5}-\sqrt[3]{4}\right)\end{cases}}\)
\(I=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-5\sqrt{3}.\sqrt{3^2}+2\sqrt{2^2}.\sqrt{3}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}\)
\(=-5\sqrt{3}.\dfrac{1}{\sqrt{3}}\)
\(=-5\)
\(K=\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}\)
\(=\sqrt{5^2.5}-4\sqrt{3^2.5}+3\sqrt{2^2.5}-\sqrt{4^2.5}\)
\(=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}\)
\(=\sqrt{5}.\left(5-12+6-4\right)\)
\(=-5\sqrt{5}\)
\(L=2\sqrt{9}+\sqrt{25}-5\sqrt{4}\)
\(=2\sqrt{3^2}+\sqrt{5^2}-5\sqrt{2^2}\)
\(=2.3+5-5.2\)
\(=1\)
\(N=2\sqrt{32}-5\sqrt{27}-4\sqrt{8}+3\sqrt{75}\)
\(=2.4\sqrt{2}-5.3\sqrt{3}-4.2\sqrt{2}+3.5\sqrt{3}\)
\(=8\sqrt{2}-8\sqrt{2}-15\sqrt{3}+15\sqrt{3}\)
\(=0\)
\(O=2\sqrt{3.5^2}-3\sqrt{3.2^2}+\sqrt{3.3^2}\)
\(=2.5\sqrt{3}-3.2\sqrt{3}+3\sqrt{3}\)
\(=10\sqrt{3}-6\sqrt{3}+3\sqrt{3}\)
\(=7\sqrt{3}\)
\(L=\dfrac{2\sqrt{3}-15\sqrt{3}+8\sqrt{3}}{\sqrt{3}}=2-15+8=-5\)
\(K=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}\)
L=2*3+5-5*2=5-4=1
N=8căn 2-8căn2-15căn3+15căn 3=0
O=10căn 3-6căn3+3căn3=7căn 3
a) Ta có: \(4\sqrt{28}+3\sqrt{63}-3\sqrt{112}-2\sqrt{175}\)
\(=8\sqrt{7}+9\sqrt{7}-12\sqrt{7}-10\sqrt{7}\)
\(=-5\sqrt{7}\)
b) Ta có: \(\sqrt{5}\left(\sqrt{5}-3\sqrt{20}+2\sqrt{80}\right)\)
\(=\sqrt{5}\left(\sqrt{5}-6\sqrt{5}+8\sqrt{5}\right)\)
\(=\sqrt{5}\cdot3\sqrt{5}=15\)
c) Ta có: \(\left(\sqrt{\dfrac{16}{3}}-\sqrt{\dfrac{25}{3}}\right)\cdot\sqrt{3}\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\sqrt{3}\)
=-1
e) Ta có: \(\left(\sqrt{\dfrac{32}{3}}-\sqrt{54}+\sqrt{\dfrac{50}{3}}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{4\sqrt{2}}{\sqrt{3}}+\dfrac{5\sqrt{2}}{\sqrt{3}}-3\sqrt{6}\right)\cdot\sqrt{6}\)
\(=\dfrac{9\sqrt{12}}{\sqrt{3}}-18\)
\(=0\)
f) Ta có: \(\left(\sqrt{6}-2\right)\left(\sqrt{3}+\sqrt{2}\right)\)
\(=3\sqrt{2}+2\sqrt{3}-2\sqrt{2}-2\sqrt{2}\)
\(=\sqrt{2}\)
a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
a: ĐKXĐ: x-5>=0
=>x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>x-5=4
=>x=9(nhận)
b: ĐKXĐ: x-1>=0
=>x>=1
\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)
=>\(\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=4\)
=>\(-2\sqrt{x-1}=4\)
=>\(\sqrt{x-1}=-2\)(vô lý)
Vậy: Phương trình vô nghiệm
c: ĐKXĐ: x-2>=0
=>x>=2
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot\sqrt{9x-18}+6\cdot\sqrt{\dfrac{x-2}{81}}=-4\)
=>\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
=>\(\sqrt{x-2}\left(\dfrac{1}{3}-2+\dfrac{2}{3}\right)=-4\)
=>\(-\sqrt{x-2}=-4\)
=>x-2=16
=>x=18(nhận)
d: ĐKXĐ: x+3>=0
=>x>=-3
\(\sqrt{9x+27}+4\sqrt{x+3}-\dfrac{3}{4}\cdot\sqrt{16x+48}=0\)
=>\(3\sqrt{x+3}+4\sqrt{x+3}-\dfrac{3}{4}\cdot4\sqrt{x+3}=0\)
=>\(4\sqrt{x+3}=0\)
=>x+3=0
=>x=-3(nhận)
a) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
= \(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)
= \(2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
= \(2\sqrt{x-5}=4\)
= \(\sqrt{x-5}=2\)
= \(\left|x-5\right|=4\)
=> \(x-5=\pm4\)
\(x=\pm4+5\)
\(x=9;x=1\)
Vậy x=9; x=1
máy tính sinh ra là để sử dụng trong các trường hợp này :)
\(\left(\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25}\right)^2\)
\(=\sqrt[3]{4}+2\sqrt[3]{50}+5\sqrt[3]{5}+2\left(2\sqrt[3]{5}-\sqrt[3]{50}-5\sqrt[3]{4}\right)\)
\(=9\sqrt[3]{5}-9\sqrt[3]{4}=9\left(\sqrt[3]{5}-\sqrt[3]{4}\right)\)
\(\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25}=3\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}\)