Cho \(b=\sqrt[3]{2020}\). Tính:
\(Q=\sqrt[3]{\dfrac{b^3-3b+\left(b^2-1\right)\sqrt{b^2-4}}{2}}+\sqrt[3]{\dfrac{b^3-3b-\left(b^2-1\right)\sqrt{b^2-4}}{2}}\)
cho b=\(\sqrt[3]{2010}\). tính A=\(\sqrt[3]{\dfrac{b^3-3b+\left(b^2-1\right)\sqrt{b^2-4}}{2}}+\sqrt[3]{\dfrac{b^3-3b-\left(b^2-1\right)\sqrt{b^2-4}}{2}}\)
Cho \(b=\sqrt[3]{2020}\). Tính \(Q=\sqrt[3]{\frac{b^3-3b+\left(b^2-1\right)\sqrt{b^2-4}}{2}}+\sqrt[3]{\frac{b^3-3b-\left(b^2-1\right)\sqrt{b^2-4}}{2}}\)
\(Q=\sqrt[3]{\frac{b^3-3b+\left(b^2-1\right)\sqrt{b^2-4}}{2}}+\sqrt[3]{\frac{b^3-3b-\left(b^2-1\right)\sqrt{b^2-4}}{2}}\)
\(\Leftrightarrow Q^3=b^3-3b+3Q\sqrt[3]{\frac{b^3-3b+\left(b^2-1\right)\sqrt{b^2-4}}{2}}.\sqrt[3]{\frac{b^3-3b-\left(b^2-1\right)\sqrt{b^2-4}}{2}}\)
\(\Leftrightarrow Q^3=b^3-3b+3Q\)
\(\Leftrightarrow\left(Q-b\right)\left(Q^2+Qb+b^2-3\right)=0\)
Dễ thấy \(Q^2+Qb+b^2-3>0\)
\(\Rightarrow Q=b=\sqrt[3]{2020}\)
a, \(A=\left(\sqrt{2}+1\right)[\left(\sqrt{2}\right)^2+1][(\sqrt{2})^4+1][\left(\sqrt{2}\right)^8+1][1\left(\sqrt{2}\right)^{16}+1]\)
b, \(B=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+1\sqrt{2020}}\)
c,\(C=^3\sqrt[]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\)
a) Ta có: \(A=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^2+1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^2+1\right]\left[\left(\sqrt{2}\right)^2-1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^4-1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^8-1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^{16}-1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^{32}-1\right]\)
\(=65535\sqrt{2}+65535\)
b) Ta có: \(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2020}}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2020}-\sqrt{2019}\)
\(=\sqrt{2020}-1\)
\(=2\sqrt{505}-1\)
c) Ta có: \(C^3=26+15\sqrt{3}+26-15\sqrt{3}+3\cdot\sqrt[3]{\left(26+15\sqrt{3}\right)\left(26-15\sqrt{3}\right)}\cdot\left(\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\right)\)
\(\Leftrightarrow C^3=52+3\cdot C\)
\(\Leftrightarrow C^3-3\cdot C-52=0\)
\(\Leftrightarrow C^3-4C^2+4C^2-16C+13C-52=0\)
\(\Leftrightarrow C^2\left(C-4\right)+4C\left(C-4\right)+13\left(C-4\right)=0\)
\(\Leftrightarrow\left(C-4\right)\left(C^2+4C+13\right)=0\)
mà \(C^2+4C+13>0\)
nên C-4=0
hay C=4
P=\(\dfrac{\left(\sqrt{a^2+a\sqrt{a^2-b^2}}-\sqrt{a^2-a\sqrt{a^2-b^2}}\right)^2}{2\sqrt{a^3b}}:\left(\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b}{a}}-2\right)\left(a>0,b>0\right)\)
a)rút gọn P
b)tính P biết a=7+\(4\sqrt{3}\) , b=7-\(4\sqrt{3}\)
a: \(G=\left(\sqrt{a^2+a\sqrt{a^2-b^2}}-\sqrt{a^2-a\sqrt{a^2-b^2}}\right)^2\)
\(=a^2+a\sqrt{a^2-b^2}+a^2-a\sqrt{a^2-b^2}-2\cdot\sqrt{a^4-a^2\left(a^2-b^2\right)}\)
\(=2a^2-2\cdot\sqrt{a^4-a^4+a^2b^2}=2a^2-2ab\)
\(A=\left(\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b}{a}}-2\right)\)
\(=\dfrac{a}{\sqrt{ab}}+\dfrac{b}{\sqrt{ab}}-\dfrac{2\sqrt{ab}}{\sqrt{ab}}=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{ab}}\)
\(P=\dfrac{2a^2-2ab}{2a\sqrt{ab}}:\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{ab}}\)
\(=\dfrac{2a\left(a-b\right)}{2a\sqrt{ab}}\cdot\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}=\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
b: Khi \(a=7+4\sqrt{3};b=7-4\sqrt{3}\) thì
\(P=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{2+\sqrt{3}-2+\sqrt{3}}=\dfrac{4}{2\sqrt{3}}=\dfrac{2}{\sqrt{3}}\)
Tính giá trị biểu thức:
a) \(P=\left(x^3+12x-9\right)^{2005}\), biết \(x=\sqrt[3]{4\left(\sqrt{5}+1\right)}-\sqrt[3]{4\left(\sqrt{5}-1\right)}\);
b) \(Q=x^3+ax+b\), biết \(x=\sqrt[3]{-\dfrac{b}{2}+\sqrt{\dfrac{b^2}{4}+\dfrac{a^3}{27}}}+\sqrt[3]{-\dfrac{b}{2}-\sqrt{\dfrac{b^2}{4}+\dfrac{a^3}{27}}}\)
a) Để tính giá trị của biểu thức P=(x^3+12x−9)^{2005}=(√3+12√−9)^{2005} với x=3√4(√5+1)−3√4(√5−1). Đầu tiên, ta thay x bằng giá trị đã cho vào biểu thức P: P=(3√4(√5+1)−3√4(√5−1))^3+12(3√4(√5+1)−3√4(√5−1))−9)^{2005} Tiếp theo, ta thực hiện các phép tính để đơn giản hóa biểu thức: P=(4(5+1)^{1/2}−4(5−1)^{1/2})^3+12(4(5+1)^{1/2}−4(5−1)^{1/2})−9)^{2005} =(4√6−4√4)^3+12(4√6−4√4)−9)^{2005} =(4√6−8)^3+12(4√6−8)−9)^{2005} =(64√6−192+96√6−96−9)^{2005} =(160√6−297)^{2005} ≈ 1.332 × 10^3975
b) Để tính giá trị của biểu thức Q=x^3+ax+b=√3+√a+√b^2+√a^3+√3+√a−√b^2+√a^3 với x=3√−b^2+√b^2/4+a^3/(27+3√−b^2−√b^2/4+a^3/27). Tương tự như trên, ta thay x bằng giá trị đã cho vào biểu thức Q: Q=(3√−b^2+√b^2/4+a^3/(27+3√−b^2−√b^2/4+a^3/27))^3+a(3√−b^2+√b^2/4+a^3/(27+3√−b^2−√b^2/4+a^3/27))+b Tiếp theo, ta thực hiện các phép tính để đơn giản hóa biểu thức: Q=(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))^3+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b =−b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b =−b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b =−b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b =−b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b ≈ −b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b
Cho a và b là các số dương. Đơn giản các biểu thức sau :
a) \(\dfrac{a^{\dfrac{4}{3}}\left(a^{-\dfrac{1}{3}}+a^{\dfrac{2}{3}}\right)}{a^{\dfrac{1}{4}}\left(a^{\dfrac{3}{4}}+a^{-\dfrac{1}{4}}\right)}\)
b) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\)
c) \(\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\left(a^{\dfrac{2}{3}}+b^{\dfrac{2}{3}}-\sqrt[3]{ab}\right)\)
d) \(\left(a^{\dfrac{1}{3}}+b^{\dfrac{1}{3}}\right):\left(2+\sqrt[3]{\dfrac{a}{b}}+\sqrt[3]{\dfrac{b}{a}}\right)\)
a)
\(A=\dfrac{a^{\dfrac{4}{3}}\left(a^{-\dfrac{1}{3}}+a^{\dfrac{2}{3}}\right)}{a^{\dfrac{1}{4}}\left(a^{\dfrac{3}{4}}+a^{-\dfrac{1}{4}}\right)}=\dfrac{a^{\left(\dfrac{4}{3}-\dfrac{1}{3}\right)+}a^{\left(\dfrac{4}{3}+\dfrac{2}{3}\right)}}{a^{\left(\dfrac{1}{4}+\dfrac{3}{4}\right)}+a^{\left(\dfrac{1}{4}-\dfrac{1}{4}\right)}}=\dfrac{a+a^2}{a+1}=\dfrac{a\left(a+1\right)}{a+1}\)
\(a>0\Rightarrow a+1\ne0\) \(\Rightarrow A=a\)
Rút gọn pt
a, \(-\dfrac{2}{3}\sqrt{\dfrac{\left(a-b\right)^3.b^5}{c}.\dfrac{9}{4}\sqrt{\dfrac{c^3}{2\left(a-b\right)}}\sqrt{ }98b}\)
b, \(\left(\sqrt{ab}+2\sqrt{\dfrac{b}{a}}-\sqrt{\dfrac{a}{b}+\dfrac{1}{ab}}\right).\sqrt{ab}\)
c, \(\left(\sqrt{b}-3\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{8}\right).2\sqrt{6}\)
d, \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
b: \(=\left(\sqrt{ab}+\dfrac{2\sqrt{ab}}{a}-\sqrt{\dfrac{a^2+1}{ab}}\right)\cdot\sqrt{ab}\)
\(=ab+\dfrac{2ab}{a}-\sqrt{a^2+1}=ab+2b-\sqrt{a^2+1}\)
c: \(=2\sqrt{6b}-6\sqrt{18}+10\sqrt{12}-\sqrt{48}\)
\(=2\sqrt{6b}-18\sqrt{2}+20\sqrt{3}-4\sqrt{3}\)
\(=2\sqrt{6n}-18\sqrt{2}+16\sqrt{3}\)
d: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)
rút gọn biểu thức
\(G=\dfrac{\sqrt[3]{a}.a^{\dfrac{2}{3}}}{\left(a^{4-2\sqrt{3}}\right)^{4+2\sqrt{3}}}\)
\(G=\dfrac{a^{\sqrt{7}+1}.a^{2-\sqrt{7}}}{\left(a^{\sqrt{2}-2}\right)^{\sqrt{2}+2}}\)
\(H=\dfrac{a^2.\left(a^{-2}.b^3\right).b^{-1}}{\left(a^{-1}.b\right)^3.a^{-5}.b^{-2}}\)
\(H=\dfrac{b^3.a^{-4}.\left(ab^2\right)^3}{\left(a^2\right)^{-2}.\left(ab^3\right)^2.b^2}\)
\(H=\dfrac{b^3.a^{-4}.\left(ab^2\right)^3}{\left(a^2\right)^{-2}.\left(ab^3\right)^2.b^2}\)
\(H=\dfrac{b^3.a^{-4}.\left(ab^2\right)^3}{\left(a^2\right)^{-2}.\left(ab^3\right)^2.b^2}\)
Bài 1: Rút gọn
a) \(\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right).\dfrac{x+2\sqrt{x}}{\sqrt{x}}\) với x>0 x≠4
b)\(\left(2+\dfrac{3+\sqrt{3}}{\sqrt{3}-1}\right).\left(2-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\)
c)\(\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
Bài 2: Cho P=\(\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\) với a>0, a≠1, a≠2
a)Rút gọn P
b)Tìm a ∈ Z để P có giá trị nguyên
Bài 1:
a)Với x > 0;x ≠ 4 ta có:
\(\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right)\cdot\dfrac{x+2\sqrt{x}}{\sqrt{x}}\)
\(=\left(\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\left(\sqrt{x}+2\right)-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\cdot\left(\sqrt{x}+2\right)\)
\(=\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}+2\right)-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4}{x-4}\)
c)\(\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right)\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\dfrac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)=b-a\)
Bài 2:
a)Với a > 0;a ≠ 1;a ≠ 2 ta có
\(P=\left(\dfrac{\sqrt{a}^3-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}^3+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\left(\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}=\dfrac{2\left(a-2\right)}{a+2}\)
b)Ta có:
\(P=\dfrac{2\left(a-2\right)}{a+2}=\dfrac{2a-4}{a+2}=\dfrac{2\left(a+2\right)-8}{a+2}=2-\dfrac{8}{a+2}\)
P nguyên khi \(2-\dfrac{8}{a+2}\) nguyên⇒\(\dfrac{8}{a+2}\) nguyên⇒\(a+2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(TH1:a+2=1\Rightarrow a=-1\left(loai\right)\)
\(TH2:a+2=-1\Rightarrow a=-3\left(loai\right)\)
\(TH3:a+2=2\Rightarrow a=0\left(loai\right)\)
\(TH4:a+2=-2\Rightarrow a=-4\left(loai\right)\)
\(TH5:a+2=4\Rightarrow a=2\left(loai\right)\)
\(TH6:a+2=-4\Rightarrow a=-6\left(loai\right)\)
\(TH7:a+2=8\Rightarrow a=6\left(tm\right)\)
\(TH8:a+2=-8\Rightarrow a=-10\left(loai\right)\)
Vậy a = 6