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ai nay dung kinh nghiem la chinh
cau a)
ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)
\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)
\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)
\(x=\frac{3-1}{1}=2\)
suy ra
x^3-4x+1=1
A=1^2018
A=1
b)
ta thay
\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)
khi do
\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)
\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)
x=2
thay vao
x^3+3x-14=0
B=0^2018
B=0
a: \(=\left(2\sqrt{7}+\sqrt{7}+2\sqrt{14}\right)\cdot\sqrt{7}-\left(51+14\sqrt{2}\right)\)
\(=3\sqrt{7}\cdot\sqrt{7}+2\sqrt{14}\cdot\sqrt{7}-51-14\sqrt{2}\)
\(=21-51=-30\)
b: \(=\dfrac{\sqrt{10}}{2}+\dfrac{\sqrt{10}-\sqrt{6}}{2}=\dfrac{2\sqrt{10}-\sqrt{6}}{2}\)
c: \(=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\sqrt{5}+\sqrt{3}}+\dfrac{\left(\sqrt{5}-\sqrt{2}\right)^2}{\sqrt{5}-\sqrt{2}}\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{2}\)
\(=2\sqrt{5}+\sqrt{3}-\sqrt{2}\)
a) \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\frac{3}{7}}\)
b) \(\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}=\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}=\frac{\left(2\sqrt{5}-\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)}=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
c) \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}=\frac{\sqrt{x}}{\sqrt{y}}\) (Bạn tự thêm đk)
d) \(\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}-1}\) (Bạn tự thêm đk)
a) \(\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\(=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}-1}\)
b) \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}=\sqrt{\frac{x}{y}}\)
c) \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\frac{3}{7}}\)
d) \(\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}=\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(\sqrt{1}-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\)
\(=\frac{\left(2\sqrt{5}-\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)}=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
e) \(\frac{-3\sqrt{3}+3}{2\sqrt{3}-2}=\frac{-3\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}=-\frac{3}{2}\)
a) \(\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)
\(=\frac{3\left(2-\sqrt{3}\right)}{2^2-3}+\frac{13\left(4+\sqrt{3}\right)}{4^2-3}+\frac{6\sqrt{3}}{3}\)
\(=3\left(2-\sqrt{3}\right)+\left(4+\sqrt{3}\right)+2\sqrt{3}\)
\(=3.2+4=6+4=10\)
b) \(=\left[\frac{\left(\sqrt{14}-\sqrt{7}\right)\left(\sqrt{2}+1\right)}{2-1}+\frac{\left(\sqrt{15}-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{3-1}\right]:\frac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\) (nhân bung mấy cái trong ngoặc vuông ra, rút gọn)
c) Gợi ý: \(28-10\sqrt{3}=5^2-2.5.\sqrt{3}+\sqrt{3}=\left(5-\sqrt{3}\right)^2\)
d) \(=\frac{3\left(3-2\sqrt{3}\right)}{3^2-\left(2\sqrt{3}\right)^2}+\frac{3\left(3+2\sqrt{3}\right)}{3^2-\left(2\sqrt{3}\right)^2}=-6\)
e) Tự làm.
g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
