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\(=\sqrt{\left(2-\sqrt{3}\right)^2\left(26+15\sqrt{3}\right)}-\sqrt{\left(2+\sqrt{3}\right)^2\left(26-15\sqrt{3}\right)}=\)
\(=\sqrt{\left(7-4\sqrt{3}\right)\left(26+15\sqrt{3}\right)}-\sqrt{\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)=}\)
\(=\sqrt{7.26+7.15\sqrt{3}-4.26\sqrt{3}-180}-\sqrt{7.26-7.15\sqrt{3}+4.26\sqrt{3}-180}=\)
\(=\sqrt{4+\sqrt{3}}-\sqrt{4-\sqrt{3}}\)
a) Ta có: \(M=\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)
\(=\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-2\sqrt{7}+3\sqrt{6}\)
\(=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)
\(=5\sqrt{6}\)
b) Ta có: \(N=\left(2-\sqrt{3}\right)\left(\sqrt{26+15\sqrt{3}}\right)-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{\left(3\sqrt{3}+5\right)^2}-\left(2+\sqrt{3}\right)\sqrt{\left(3\sqrt{3}-5\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\)
\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
Xét: \(A=\sqrt{26+15\sqrt{3}}\) dễ thấy A > 0
\(\Leftrightarrow A^2=52-2\sqrt{26^2-15^2.3}=50\Leftrightarrow A=\sqrt{50}\)
Vậy: \(A=2+\sqrt{3}.\sqrt{26+15\sqrt{3}}-2\sqrt{3}.\sqrt{26-15\sqrt{3}}\)
\(=2+\sqrt{3}.A=2+\sqrt{3}.\sqrt{50}=5\sqrt{6}+10\sqrt{2}\)
\(\sqrt[3]{53\sqrt{5}+124}+\sqrt[3]{32\sqrt{5}-72}\)
\(=\sqrt[3]{\left(\sqrt{5}\right)^3+3.5.4+3.\sqrt{5}.4+4^3}+\sqrt[3]{\left(\sqrt{5}\right)^3-3.5.3+3.\sqrt{5}.3^2-3^3}\)
\(=\sqrt[3]{\left(\sqrt{5}+4\right)^3}+\sqrt[3]{\left(\sqrt{5}-3\right)^3}\)
\(=\sqrt{5}+4+\sqrt{5}-3\)
\(=2\sqrt{5}+1\)
Ý tưởng; Dựa vào "thế hình" của đề bài, ta dự đoán biểu thức trong căn sẽ phân tích được thành lũy thừa bậc 3 của một biểu thức có dạng \(a+\sqrt{3}\left(a\inℤ\right)\). Ta thấy:
\(\left(a+\sqrt{3}\right)^3=a^3+3a^2\sqrt{3}+9a+3\sqrt{3}\) \(=a^3+9a+\left(3a^2+3\right)\sqrt{3}\)
Ta có hệ: \(\left\{{}\begin{matrix}a^3+9a=26\\3a^2+3=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\left(a^2+9\right)=26\\a^2+1=5\end{matrix}\right.\)
\(\Rightarrow a=\pm2\). Nhưng rõ ràng \(a=-2\) không thỏa. Vậy \(a=2\).
Trình bày: Ta có
\(\sqrt[3]{26+15\sqrt{3}}\)
\(=\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}\)
\(=\sqrt[3]{2^3+3.2^2.\sqrt{3}+3.2\left(\sqrt{3}\right)^2+\left(\sqrt{3}\right)^3}\)
\(=\sqrt[3]{\left(2+\sqrt{3}\right)^3}\)
\(=2+\sqrt{3}\)
22,
1, Đặt √(3-√5) = A
=> √2A=√(6-2√5)
=> √2A=√(5-2√5+1)
=> √2A=|√5 -1|
=> A=\(\dfrac{\sqrt{5}-1}{\text{√2}}\)
=> A= \(\dfrac{\sqrt{10}-\sqrt{2}}{2}\)
2, Đặt √(7+3√5) = B
=> √2B=√(14+6√5)
=> √2B=√(9+2√45+5)
=> √2B=|3+√5|
=> B= \(\dfrac{3+\sqrt{5}}{\sqrt{2}}\)
=> B= \(\dfrac{3\sqrt{2}+\sqrt{10}}{2}\)
3,
Đặt √(9+√17) - √(9-√17) -\(\sqrt{2}\)=C
=> √2C=√(18+2√17) - √(18-2√17) -\(2\)
=> √2C=√(17+2√17+1) - √(17-2√17+1) -\(2\)
=> √2C=√17+1- √17+1 -\(2\)
=> √2C=0
=> C=0
26,
|3-2x|=2\(\sqrt{5}\)
TH1: 3-2x ≥ 0 ⇔ x≤\(\dfrac{-3}{2}\)
3-2x=2\(\sqrt{5}\)
-2x=2\(\sqrt{5}\) -3
x=\(\dfrac{3-2\sqrt{5}}{2}\) (KTMĐK)
TH2: 3-2x < 0 ⇔ x>\(\dfrac{-3}{2}\)
3-2x=-2\(\sqrt{5}\)
-2x=-2√5 -3
x=\(\dfrac{3+2\sqrt{5}}{2}\) (TMĐK)
Vậy x=\(\dfrac{3+2\sqrt{5}}{2}\)
2, \(\sqrt{x^2}\)=12 ⇔ |x|=12 ⇔ x=12, -12
3, \(\sqrt{x^2-2x+1}\)=7
⇔ |x-1|=7
TH1: x-1≥0 ⇔ x≥1
x-1=7 ⇔ x=8 (TMĐK)
TH2: x-1<0 ⇔ x<1
x-1=-7 ⇔ x=-6 (TMĐK)
Vậy x=8, -6
4, \(\sqrt{\left(x-1\right)^2}\)=x+3
⇔ |x-1|=x+3
TH1: x-1≥0 ⇔ x≥1
x-1=x+3 ⇔ 0x=4 (KTM)
TH2: x-1<0 ⇔ x<1
x-1=-x-3 ⇔ 2x=-2 ⇔x=-1 (TMĐK)
Vậy x=-1
Đặt bt là A
\(A\sqrt{2}=\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt{\left(3\sqrt{3}+5\right)^2}-\left(2+\sqrt{3}\right)\sqrt{\left(3\sqrt{3}-5\right)^2}\)\(=\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)\)
khai triển hết ra ta đc \(A\sqrt{2}=2\Rightarrow A=\sqrt{2}\)
Lời giải:
Gọi biểu thức trên là $A$
Đặt \(\sqrt[3]{15\sqrt{3}-26}=a; \sqrt[3]{15\sqrt{3}+26}=b\). Ta có:
\(a^3-b^3=-52\)
\(ab=-1\)
\(A^3=(a-b)^3=a^3-3ab(a-b)-b^3=-52+3A\)
\(\Leftrightarrow A^3-3A+52=0\)
\(\Leftrightarrow A^2(A+4)-4A(A+4)+13(A+4)=0\)
\(\Leftrightarrow (A+4)(A^2-4A+13)=0\)
Dễ thấy $A^2-4A+13>0$ nên $A+4=0$
$\Leftrightarrow A=-4$