vì sao \(\sqrt{84}=2\sqrt{21}\)?
Tính:
a) \(\frac{-90}{189}+\frac{45}{84}-\frac{78}{126}\)
b) \(11\frac{7}{21}+8\frac{16}{24}-10\frac{21}{35}\)
c) \(\frac{37}{43}\cdot\frac{17}{29}-\frac{21}{41}\cdot\frac{1}{2}+\frac{9}{58}:1\frac{6}{37}-\frac{6}{2}:1\frac{29}{21}\)
d) \(\sqrt{\left(-5\right)^2}+\sqrt{5^2}-\sqrt{\left(-5\right)^2}-\sqrt{3^2}\)
a) -90/189 + 45/84 - 78/126
= -10/21 + 15/28 - 13/21
= (-10/21 - 13/21) + 15/28
= -24/21 + 15/28
= -17/28
a, A = \(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{14\sqrt{2}-20}\)
b, X = \(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
\(A=\sqrt[3]{2^3+3.2^2.\sqrt{2}+3.2.\sqrt{2}^2+\sqrt{2}^3}+\sqrt[3]{\sqrt{2}^3-3.\sqrt{2}^2.2+3.\sqrt{2}.2^2-2^3}\)
\(A=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(\sqrt{2}-2\right)^3}\)
\(A=2+\sqrt{2}+\sqrt{2}-2=2\sqrt{2}\)
\(X=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
\(\Rightarrow X^3=\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)^3\)
\(\Rightarrow X^3=2+3\sqrt[3]{1-\frac{84}{81}}\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)\)
\(\Rightarrow X^3=2-3\sqrt[3]{\frac{1}{27}}.X\)
\(\Rightarrow X^3=2-X\)
\(\Rightarrow X^3+X-2=0\)
\(\Rightarrow\left(X-1\right)\left(X^2+2X+2\right)=0\)
\(\Rightarrow X=1\) (do \(X^2+2X+2=\left(X+1\right)^2+1>0\) \(\forall X\))
a) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
b) \(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
a) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
\(=\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(=\left(3\sqrt{7}-2\sqrt{3}\right)\sqrt{7}+2\sqrt{21}\)
\(=21-2\sqrt{21}+2\sqrt{21}\)
\(=21\)
b) \(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}\cdot1+1^2}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{2}\left(\sqrt{3}-1\right)}\)
\(=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}\)
\(=\dfrac{1}{\sqrt{2}}\)
\(\left(\sqrt{7}-\sqrt{3}\right)^2+\sqrt{84}\)
\(\left(\sqrt{7}-\sqrt{3}\right)^2+\sqrt{84}\)
⇔ \(\left(\sqrt{7}\right)^2-2.\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2+\sqrt{84}\)
⇔ \(7-2\sqrt{21}+3+2\sqrt{21}\)
⇔ \(10\)
Chứng minh rằng các biểu thức sau là 1 số nguyên:
a) \(A=\sqrt[3]{20+14\sqrt{2}}-\sqrt[3]{14\sqrt{2}-20}\)
b) \(B=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
Bài 1 Tính giá trị biểu thức
a. \(\sqrt{0,16}\)+ \(\sqrt{0,04}\) - \(\sqrt{0,25}\)
b.\(\sqrt{85^2-84^2}\) - \(\sqrt{26^2-24^2}\)
a) \(\sqrt{0,16}+\sqrt{0,04}-\sqrt{0,25}\)
= 0,4 + 0,2 - 0,5
= 0,1
b) \(\sqrt{85^2-84^2}-\sqrt{26^2-24^2}\)
= \(\sqrt{\left(85-84\right)\left(85+84\right)}\) - \(\sqrt{\left(26-24\right)\left(26+24\right)}\)
= \(\sqrt{169}\) - \(\sqrt{2.50}\)
= 13 - 10
= 3
Chúc bạn học tốt
a) Ta có: \(\sqrt{0.16}+\sqrt{0.04}-\sqrt{0.25}\)
\(=0,4+0,2-0,5\)
=0,1
b) Ta có: \(\sqrt{85^2-84^2}-\sqrt{26^2-24^2}\)
=13-10
=3
\(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
\(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
\(=\sqrt{196}-\sqrt{84}+7+\sqrt{84}\)
\(=14+7=21\)
mk không chép đề
(2\(\sqrt{7}\)-2\(\sqrt{3}\)+\(\sqrt{7}\))\(\sqrt{7}\) +2\(\sqrt{21}\)
=14 -2\(\sqrt{21}\)+7 +2\(\sqrt{21}\)
=21
CM: \(\sqrt[3]{1+\dfrac{\sqrt{84}}{9}}+\sqrt[3]{1-\dfrac{\sqrt{84}}{9}}\) là số nguyên
Rút gọn
a) \(\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}+2}\)
b) \(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
b: \(=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\)
\(=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\)
\(=50-10\sqrt{21}+10\sqrt{21}-42=8\)
a: \(A=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}\)
=>\(A^2=\sqrt{2}-1+\sqrt{2}+1+2\sqrt{2-1}=2\sqrt{2}+2\)
=>\(A=\sqrt{2\sqrt{2}+2}\)
Đặt \(B=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2+\sqrt{2}}\)
=>\(B=\sqrt{2\sqrt{2}+2}-\sqrt{2+\sqrt{2}}\)
=>\(B^2=2\sqrt{2}+2+2+\sqrt{2}-2\sqrt{\sqrt{2}\left(2+\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)
=>\(B^2=4+3\sqrt{2}-2\sqrt[4]{2}\left(2+\sqrt{2}\right)\)
=>\(B\simeq0,35\)