Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{1}\cdot\dfrac{\sqrt{x}-1}{2}\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
Ta có: \(x=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=1\)
Thay x=1 vào B, ta được:
\(B=-\sqrt{1}\cdot\left(\sqrt{1}-1\right)=0\)
1) \(P=\dfrac{\sqrt{x}.\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
2) \(P=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{3+2\sqrt{2}}+1\right)^2}{\sqrt{3+2\sqrt{2}}}\)
\(=\dfrac{\left(\sqrt{\left(\sqrt{2}+1\right)^2}+1\right)^2}{\sqrt{\left(\sqrt{2}+1\right)^2}}=\dfrac{\left(\sqrt{2}+2\right)^2}{\sqrt{2}+1}=\dfrac{2+4+4\sqrt{2}}{\sqrt{2}+1}=\dfrac{6+4\sqrt{2}}{\sqrt{2}+1}\)
\(=>x^3=(\sqrt[3]{2\left(\sqrt{3}+1\right)}-\sqrt[3]{2\left(\sqrt{3}-1\right)})^3\)
\(x^3=2\left(\sqrt{3}+1\right)-3.\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]^2.\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\)
+\(3\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]^2\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]-2\left(\sqrt{3}-1\right)\)
\(x^3=\)
\(4-3\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}-\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\)
\(x^3=4-3.\left[\sqrt[3]{4\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\right].\)\(x\)
\(x^3=4-3\left[\sqrt[3]{4\left(3-1\right)}\right].x\)
\(x^3=4-3.2x\)
\(x^3=4-6x\)
thay \(x^3=4-6x\) vào A=>\(A=\left(4-6x+6x-5\right)^{2009}=\left(-1\right)^{2009}=-1\)
Bài 1:
a: \(=\left|5-\sqrt{3}\right|-\left|\sqrt{3}-2\right|\)
\(=5-\sqrt{3}-2+\sqrt{3}=3\)
b; \(B=\dfrac{\left(2-\sqrt{3}\right)\cdot\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\cdot\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\cdot\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-6\sqrt{3}+10-9+5\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{20-18}{\sqrt{2}}=\sqrt{2}\)
c: \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3+3-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}=1\)
d: \(A=\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)
\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)
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
\(a.\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)\sqrt{2}\sqrt{2+\sqrt{3}}.\)
\(=\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}=\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3+1}\right)^2}\)
\(=\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)^2=\left(2-\sqrt{3}\right)\left(4+2\sqrt{3}\right)\)
\(=2\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2\left(2^2-\sqrt{3}^2\right)=2\)
\(1.A=x-3\sqrt{x}+5=\left(\sqrt{x}-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\) Điều kiện: \(x\ge0\)
\(\Rightarrow MinA=\frac{11}{4}\)
Dấu "=" xảy ra khi \(\sqrt{x}=\frac{3}{2}\Leftrightarrow x=\frac{9}{4}\left(TM\right)\)
\(2.B=\left(x-2015\right)-\sqrt{x-2015}+2015=\left(\sqrt{x-2015}-\frac{1}{2}\right)^2+2015-\frac{1}{4}\) điều kiện: \(x\ge2015\)
\(B\ge2015-\frac{1}{4}=\frac{8059}{8060}\)
Dấu "=" xảy ra khi \(\sqrt{x-2015}-\frac{1}{2}=0\Leftrightarrow x-2015=\frac{1}{2^2}\Leftrightarrow x=\frac{8061}{8060}\left(TM\right)\)
\(A=\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\dfrac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(x=\dfrac{9-4\sqrt{5}-9-4\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}:2\sqrt{5}=\dfrac{-8\sqrt{5}}{-2\sqrt{5}}=4\\ \Leftrightarrow\sqrt{x}=2\\ \Leftrightarrow A=\dfrac{2-1}{2+2}=\dfrac{1}{4}\)
\(x=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
=1
Thay x=1 vào B, ta được:
\(B=-\sqrt{1}\cdot\left(\sqrt{1}-1\right)=0\)