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CD
0
20 tháng 7 2015
đặt \(a=\sqrt[3]{6+\sqrt{\frac{847}{27}}};b=\sqrt[3]{6-\sqrt{\frac{847}{27}}}\). dễ thấy a> 0; b > 0
=> \(a^3+b^3=6+\sqrt{\frac{847}{27}}+6-\sqrt{\frac{847}{27}}=12\); \(a.b=\sqrt[3]{6+\sqrt{\frac{847}{27}}}.\sqrt[3]{6-\sqrt{\frac{847}{27}}}=\sqrt[3]{36-\frac{847}{27}}=\frac{5}{3}\)
Có: (a+ b)3 = a3 + b3 + 3ab (a+ b)
=> (a + b)3 = 12 + 3. \(\frac{5}{3}\).(a + b) = 12+ 5.(a + b)
=> (a + b)3 - 5.(a +b) - 12 = 0
<=> (a + b)3 - 9.(a + b) + 4.(a + b) - 12 = 0
<=> (a + b). [(a + b)2 - 9] + 4.(a + b - 3) = 0 <=> (a + b).(a + b + 3).(a + b- 3) + 4.(a + b - 3) = 0
<=> (a+ b - 3).[(a + b)(a+ b+ 3) + 4] = 0
<=> a+ b = 3 hoặc (a + b)(a+ b+ 3) + 4 = 0
tuy nhiên : Vì a > 0; b > 0 nên (a + b)(a+ b+ 3) + 4 > 0
vậy a + b = 3 => điều phải chứng minh
NN
8 tháng 9 2020
2. a) \(ĐKXĐ:x\ge\frac{1}{3}\)
\(\sqrt{3x-1}=4\)\(\Rightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)\(\Leftrightarrow3x=17\)\(\Leftrightarrow x=\frac{17}{3}\)( thỏa mãn ĐKXĐ )
Vậy \(x=\frac{17}{3}\)
b) \(ĐKXĐ:x\ge1\)
\(\sqrt{x-1}=x-1\)\(\Rightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)( thỏa mãn ĐKXĐ )
Vậy \(x=1\)hoặc \(x=2\)
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
Vì \(6>1\)\(\Leftrightarrow\sqrt{6}>\sqrt{1}=1\)\(\Rightarrow\sqrt{6}-1>0\)
\(6>4\)\(\Rightarrow\sqrt{6}>\sqrt{4}=2\)\(\Rightarrow\sqrt{6}-2>0\)
\(\Rightarrow\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|=\left(\sqrt{6}-1\right)-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2=1\)
hay \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=1\)
8 tháng 9 2020
2a) \(\sqrt{3x-1}=4\)( ĐKXĐ : \(x\ge\frac{1}{3}\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)
\(\Leftrightarrow3x=17\)
\(\Leftrightarrow x=\frac{17}{3}\)( tmđk )
Vậy phương trình có nghiệm duy nhất là x = 17/3
b) \(\sqrt{x-1}=x-1\)( ĐKXĐ : \(x\ge1\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)
\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}\left(tmđk\right)}\)
Vậy phương trình có hai nghiệm là x = 1 hoặc x = 2
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)
\(=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}-\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot2+2^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}\)
\(=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
\(=\sqrt{6}-1-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2\)
\(=1\)
AH
Akai Haruma
Giáo viên
30 tháng 8 2019
a)
\(\sqrt{12}-\sqrt{27}+\sqrt{3}=\sqrt{4}.\sqrt{3}-\sqrt{9}.\sqrt{3}+\sqrt{3}=2\sqrt{3}-3\sqrt{3}+\sqrt{3}\)
\(=\sqrt{3}(2-3+1)=0\)
b)
\(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}=\sqrt{4}.\sqrt{63}-\sqrt{4}.\sqrt{175}+\sqrt{4}.\sqrt{252}-\sqrt{4}.\sqrt{112}\)
\(=2(\sqrt{63}-\sqrt{175}+\sqrt{252}-\sqrt{112})\)
\(=2(\sqrt{9}.\sqrt{7}-\sqrt{25}.\sqrt{7}+\sqrt{36}.\sqrt{7}-\sqrt{16}.\sqrt{7})\)
\(=2(3\sqrt{7}-5\sqrt{7}+6\sqrt{7}-4\sqrt{7})=2\sqrt{7}(3-5+6-4)=0\)
------------------
\(\sqrt{3}(\sqrt{12}+\sqrt{27}-\sqrt{3})=\sqrt{36}+\sqrt{81}-\sqrt{9}\)
\(=\sqrt{6^2}+\sqrt{9^2}-\sqrt{3^2}=6+9-3=12\)
AH
Akai Haruma
Giáo viên
30 tháng 8 2019
c)
\(\frac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}=\frac{\sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{5}}{\sqrt{7}.\sqrt{3}+\sqrt{7}.\sqrt{5}}=\frac{\sqrt{2}(\sqrt{3}+\sqrt{5})}{\sqrt{7}(\sqrt{3}+\sqrt{5})}=\frac{\sqrt{2}}{\sqrt{7}}\)
\(\frac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}=\frac{\sqrt{81}.\sqrt{5}+3\sqrt{9}.\sqrt{3}}{3\sqrt{3}+\sqrt{9}.\sqrt{5}}=\frac{9\sqrt{5}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}\)
\(=\frac{3(3\sqrt{5}+3\sqrt{3})}{3\sqrt{3}+3\sqrt{5}}=3\)
d)
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-(\sqrt{6}+\sqrt{9}+\sqrt{12})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-(\sqrt{2}.\sqrt{3}+\sqrt{3}.\sqrt{3}+\sqrt{3}.\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{3}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})(1-\sqrt{3})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1-\sqrt{3}\)
16 tháng 5 2017
Bạn không sửa thì m sửa.
Sửa đề: \(P=\sqrt[3]{\sqrt{\frac{2303}{27}}+6}-\sqrt[3]{\sqrt{\frac{2303}{27}}-6}\)
\(P^3=\sqrt{\frac{2303}{27}}+6-\left(\sqrt{\frac{2303}{27}}-6\right)-\frac{3.11.P}{3}\)
\(\Leftrightarrow P^3=12-11P\)
\(\Leftrightarrow P^3+11P-12=0\)
\(\Leftrightarrow\left(P-1\right)\left(P^2+P+12\right)=0\)
Vì \(P^2+P+12>0\) nên ta có
\(P=1\)
21 tháng 7 2020
Bài 1:
a) Ta có: \(\sqrt{243}-\frac{1}{2}\sqrt{12}-2\sqrt{75}+\sqrt{27}\)
\(=\sqrt{3}\cdot9-\frac{1}{2}\cdot\sqrt{3}\cdot2-2\cdot\sqrt{3}\cdot5+\sqrt{3}\cdot3\)
\(=\sqrt{3}\left(9-1-10+3\right)\)
\(=\sqrt{3}\cdot1=\sqrt{3}\)
b) Ta có: \(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{5}{1+\sqrt{6}}-6\sqrt{\frac{1}{6}}\)
\(=\frac{\left(2\sqrt{3}-3\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(\sqrt{3}+\sqrt{2}\right)}+\frac{5\cdot\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\sqrt{36\cdot\frac{1}{6}}\)
\(=-\sqrt{6}+\frac{5\left(\sqrt{6}-1\right)}{5}-\sqrt{6}\)
\(=-2\sqrt{6}+\sqrt{6}-1\)
\(=-\sqrt{6}-1\)
Bài 2: Rút gọn
Ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
hình thức đăng vui phương pháp lập phương hai vế sau đó nhẩm nghiệm dùng tiếp sơ đồ hoc-ne :))) là ok
\(x^3=6+\sqrt{\frac{847}{27}}+6-\sqrt{\frac{847}{27}}+3.\sqrt[3]{\left[6^2-\left(\sqrt{\frac{847}{27}}\right)^2\right]}.x\)
\(\Rightarrow x^3=12+3.\sqrt[3]{\frac{125}{27}}x\)
\(\Leftrightarrow x^3-5x-12=0\)
\(\Leftrightarrow x^3-9x+4x-12=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+4\right)=0\).Vì \(x^2+3x+4=x^2+2.\frac{3}{2}.x+\frac{9}{4}+\frac{7}{4}=\left(x+\frac{3}{2}\right)^2+\frac{7}{4}>0\)
\(\Leftrightarrow x=3\)