Bài 1 a) \(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
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Bài 1:
a) \(\sqrt{1,44\cdot1,21-1,44\cdot0,4}\)
\(=\sqrt{1,44\cdot\left(1,21-0,4\right)}\)
\(=\sqrt{1,44\cdot0,81}\)
\(=\sqrt{1,44}\cdot\sqrt{0,81}\)
\(=1,2\cdot0,9\)
\(=1,08\)
b) \(\dfrac{\sqrt{5}-2}{\sqrt{5}+2}+\sqrt{80}\)
\(=\dfrac{\left(\sqrt{5}-2\right)^2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+4\sqrt{5}\)
\(=\dfrac{5-4\sqrt{5}+4}{1}+4\sqrt{5}\)
\(=9-4\sqrt{5}+4\sqrt{5}\)
\(=9\)
c) \(\sqrt[3]{16}+\sqrt[3]{2}\left(\sqrt[3]{4}-\sqrt[3]{2}\right)\)
\(=\sqrt[3]{2^3\cdot2}+\sqrt[3]{2\cdot4}-\sqrt[3]{2\cdot2}\)
\(=2\sqrt[3]{2}+\sqrt[3]{8}-\sqrt[3]{4}\)
\(=2\sqrt[3]{2}+2-\sqrt[3]{4}\)
Bài 2: Ta có:
\(VT=\dfrac{1}{\sqrt{a}-\sqrt{b}}:\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}:\dfrac{\sqrt{ab}\cdot\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\cdot\dfrac{1}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{1}{a-b}=VP\left(dpcm\right)\)
Trục căn thức lên rồi tính như bình thường thôi.
Bạn chưa hiểu thì nhắn tin cho mình, mình làm cho.
Chúc bạn học tốt nhea. =)
a) \(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
=\(\frac{\left(\sqrt{5}+\sqrt{3}\right)^2+\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)
=\(\frac{8+2\sqrt{15}+8-2\sqrt{15}}{2}\)
=\(\frac{16}{2}=8\)
Tạm thời mih bận nên chỉ kịp lm 1 câu thôi, khi nao có time mih lm típ nha
`a)A=(3-sqrt5)sqrt{3+sqrt5}+(3+sqrt5)sqrt{3-sqrt5}`
`=sqrt{3-sqrt5}sqrt{3+sqrt5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=sqrt{9-5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=2(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=sqrt2(sqrt{6+2sqrt5}+sqrt{6-2sqrt5})`
`=sqrt2(sqrt{(sqrt5+1)^2}+sqrt{(sqrt5+1)^2})`
`=sqrt2(sqrt5+1+sqrt5-1)`
`=sqrt{2}.2sqrt5`
`=2sqrt{10}`
`b)B=(5+sqrt{21})(sqrt{14}-sqrt6)sqrt{5-sqrt{21}}`
`=sqrt{5+sqrt{21}}sqrt{5-sqrt{21}}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=sqrt{25-21}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=2sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=2sqrt2sqrt{5+sqrt{21}}(sqrt{7}-sqrt3)`
`=2sqrt{10+2sqrt{21}}(sqrt{7}-sqrt3)`
`=2sqrt{(sqrt3+sqrt7)^2}(sqrt{7}-sqrt3)`
`=2(sqrt3+sqrt7)(sqrt{7}-sqrt3)`
`=2(7-3)`
`=8`
`c)C=sqrt{4+sqrt7}-sqrt{4-sqrt7}`
`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`
`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7+1)^2/2}`
`=(sqrt7+1)/sqrt2-(sqrt7-1)/2`
`=2/sqrt2=sqrt2`
bài 3 :
nhân đảo ngược căn 2 - căn 3 rồi quy đồng là ra ngay
b: Ta có: \(\dfrac{4}{\sqrt{3}+1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{6}{3-\sqrt{3}}\)
\(=2\sqrt{3}-2+\sqrt{3}+1-3-\sqrt{3}\)
\(=2\sqrt{3}-4\)
Bài 1:
a: \(\sqrt{27}+\dfrac{1}{2}\sqrt{48}-\sqrt{108}\)
\(=3\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-6\sqrt{3}\)
\(=-3\sqrt{3}+2\sqrt{3}=-\sqrt{3}\)
b: \(\left(\sqrt{14}-\sqrt{10}\right)\cdot\sqrt{6+\sqrt{35}}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{2}\cdot\sqrt{6+\sqrt{35}}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{12+2\sqrt{35}}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)=7-5=2\)
c: \(\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}-\dfrac{2}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\dfrac{2\left(\sqrt{3}+1\right)}{3-1}\)
\(=\sqrt{3}-\sqrt{3}-1=-1\)
Bài 2:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
\(A=\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-5+\sqrt{x}-1+2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b: A=2
=>\(\sqrt{x}=2\left(\sqrt{x}-1\right)\)
=>\(2\sqrt{x}-2=\sqrt{x}\)
=>\(\sqrt{x}=2\)
=>x=4(nhận)
c: Để A là số nguyên thì \(\sqrt{x}⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1+1⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1\inƯ\left(1\right)\)
=>\(\sqrt{x}-1\in\left\{1;-1\right\}\)
=>\(\sqrt{x}\in\left\{2;0\right\}\)
=>\(x\in\left\{4;0\right\}\)
\(A=-\dfrac{3+\sqrt{5}+3-\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\cdot\dfrac{\sqrt{5}}{5}\\ A=\dfrac{-6}{4}\cdot\dfrac{\sqrt{5}}{5}=\dfrac{-3\sqrt{5}}{10}\)
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
Bài 1:
a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-16\sqrt{3}\)
b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)
\(=3-\sqrt{6}+\sqrt{6}-1\)
=3-1=2
c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)
\(=\sqrt{15}+4-\sqrt{15}=4\)
d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=3+\sqrt{5}-\dfrac{\sqrt{5}}{2}=3+\dfrac{\sqrt{5}}{2}\)
Bài 2:
Vẽ đồ thị:
Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x-4=-3x+3\)
=>\(\dfrac{1}{2}x+3x=3+4\)
=>\(\dfrac{7}{2}x=7\)
=>x=2
Thay x=2 vào y=-3x+3, ta được:
\(y=-3\cdot2+3=-3\)
Vậy: (d1) cắt (d2) tại A(2;-3)
Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(\sqrt{2}A=\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
\(\Rightarrow A=\frac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)