Tính giá trị của các biểu thức sau:
a) A = 27 - 7|7x| - 2013x, biết |x| = \(\sqrt{\left(-2\right)^2}\)
b) B = \(\left(\dfrac{1}{2013^2}-1\right)\left(\dfrac{1}{2012^2}-1\right)\left(\dfrac{1}{2011^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)
cho \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\). hãy tính giá trị biểu thức sau: \(A=f\left(\dfrac{1}{2012}\right)+f\left(\dfrac{2}{2012}\right)+...+f\left(\dfrac{2010}{2012}\right)+f\left(\dfrac{2011}{2012}\right)\)
Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)
1/ Chứng tỏ rằng : B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{8^2}< 1\)
2/ Rút gọn: B=\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{20}\right)\)
3/ Tính giá trị của biểu thức: A= \(\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
4/ So sánh : A= \(\dfrac{2011+2012}{2010+2013}\) và B= \(\dfrac{2011}{2012}+\dfrac{2012}{2013}\)
1/ \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}\)
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{8}< 1\)
\(B< 1\)
2/ \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{19}{20}\)
\(B=\dfrac{1\times2\times3\times...\times19}{2\times3\times4\times...\times20}\)
\(B=\dfrac{1}{20}\)
3/ \(A=\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)
\(A=\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}\cdot\dfrac{4}{21}\)
\(A=11\)
4/ A phải là \(\dfrac{2011+2012}{2012+2013}\)
Ta có : \(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2013}+\dfrac{2012}{2013}=\dfrac{2011+2012}{2013}>\dfrac{2011+2012}{2012+2013}=A\)
\(\Rightarrow B>A\)
Tính giá trị của các biểu thức sau :
a)\(\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)\)+(0,4 - 5) - \(\left(4\dfrac{1}{4}-1\right)\)
b)\(\dfrac{2}{3}\) - \(\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
c)\(\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right)\):\(\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
d)3 - \(\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}\)
giúp mình nhé trả lời mình cho tick cảm ơn các bạn !
\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)
\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)
\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)
\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)
\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)
\(=7-\dfrac{26}{5}\)
\(=\dfrac{9}{5}\)
\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)
\(=\dfrac{2}{3}+\dfrac{21}{8}\)
\(=\dfrac{79}{24}\)
\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)
\(=\dfrac{31}{4}:\dfrac{49}{8}\)
\(=\dfrac{62}{49}\)
\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)
tính giá trị biểu thức sau
a) \(A=3^{\dfrac{2}{5}}.3^{\dfrac{1}{5}}.3^{\dfrac{1}{5}}\)
b) \(B=\left(-27\right)^{\dfrac{1}{3}}\)
c) \(C=\sqrt[3]{-64}.\left(\dfrac{1}{2}\right)^3\)
d) \(D=\left(-27\right)^{\dfrac{1}{3}}.\left(\dfrac{1}{3}\right)^4\)
e) \(E=\left(\sqrt{3}+1\right)^{106}.\left(\sqrt{3}-1\right)^{106}\)
f) \(F=360^{\sqrt{5}+1}.20^{3-\sqrt{5}}.18^{3-\sqrt{5}}\)
g) \(G=2023^{\left(3+2\sqrt{2}\right)}.2023^{\left(2\sqrt{2}-3\right)}\)
a: \(A=3^{\dfrac{2}{5}}\cdot3^{\dfrac{1}{5}}\cdot3^{\dfrac{1}{5}}=3^{\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{1}{5}}=3^{\dfrac{4}{5}}\)
b: \(B=\left(-27\right)^{\dfrac{1}{3}}=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}=\left(-3\right)^{\dfrac{1}{3}\cdot3}=\left(-3\right)^1=-3\)
c: \(C=\sqrt[3]{-64}\cdot\left(\dfrac{1}{2}\right)^3\)
\(=\sqrt[3]{\left(-4\right)^3}\cdot\dfrac{1}{2^3}=-4\cdot\dfrac{1}{8}=-\dfrac{4}{8}=-\dfrac{1}{2}\)
d: \(D=\left(-27\right)^{\dfrac{1}{3}}\cdot\left(\dfrac{1}{3}\right)^4\)
\(=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}\cdot\dfrac{1}{3^4}\)
\(=\left(-3\right)^{3\cdot\dfrac{1}{3}}\cdot\dfrac{1}{81}=\dfrac{-3}{81}=\dfrac{-1}{27}\)
e: \(E=\left(\sqrt{3}+1\right)^{106}\cdot\left(\sqrt{3}-1\right)^{106}\)
\(=\left[\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\right]^{106}\)
\(=\left(3-1\right)^{106}=2^{106}\)
f: \(F=360^{\sqrt{5}+1}\cdot20^{3-\sqrt{5}}\cdot18^{3-\sqrt{5}}\)
\(=360^{\sqrt{5}+1}\cdot\left(20\cdot18\right)^{3-\sqrt{5}}\)
\(=360^{\sqrt{5}+1}\cdot360^{3-\sqrt{5}}=360^{\sqrt{5}+1+3-\sqrt{5}}=360^4\)
g: \(G=2023^{3+2\sqrt{2}}\cdot2023^{2\sqrt{2}-3}\)
\(=2023^{3+2\sqrt{2}+2\sqrt{2}-3}\)
\(=2023^{4\sqrt{2}}\)
Cho biểu thức A = \(\left(\dfrac{\sqrt{x}+2}{x-1}-\dfrac{\sqrt{x}-2}{x-2\sqrt{x}+1}\right):\dfrac{4x}{\left(x-1\right)^2}\)
a) Rút gọn A.
b) tính giá trị của A biết \(\left|x-5\right|=4\).
ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right).\dfrac{\left(x-1\right)^2}{4x}\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{4x}\)
\(=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
b.
\(\left|x-5\right|=4\Rightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{\sqrt{9}+1}{2\sqrt{9}}=\dfrac{2}{3}\)
7) cho biểu thức: P=\(\left(\dfrac{2}{\sqrt{x}-1}-\dfrac{5}{x+\sqrt{x}-2}\right)\div\left(1+\dfrac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)
a) rút gọn P
b) tính P khi \(x=6-2\sqrt{5}\)
c) tính giá trị của x để P= \(\dfrac{1}{\sqrt{x}}\)
giúp mk vs ah mk cần gấp lắm
đk : \(x\ge0,x\ne1\)
\(=>P=\left[\dfrac{2\left(\sqrt{x}+2\right)-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]\)
\(P=\left[\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right].\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\right]\)
\(P=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b,\(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\) thay vào P
\(=>P=\dfrac{2\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}=\dfrac{2\sqrt{5}-3}{\sqrt{5}}\)
c,\(=>\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}=>2x-\sqrt{x}=\sqrt{x}+1\)
\(=>2x-2\sqrt{x}-1=0< =>2\left(x-\sqrt{x}-\dfrac{1}{2}\right)=0\)
\(=>x-\sqrt{x}-\dfrac{1}{2}=>\Delta=1-4\left(-\dfrac{1}{2}\right)=3>0=>\left[{}\begin{matrix}x1=\dfrac{1+\sqrt{3}}{2}\\x2=\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\)
đối chiếu đk loại x2 còn x1 thỏa
cho biểu thức :\(B=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right)\)
a) Rút gọn biểu thức B
b) Tính giá trị của B khi x=\(4+2\sqrt{3}\)
a) \(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)
\(\Leftrightarrow B=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow B=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(\Leftrightarrow B=\dfrac{2-\sqrt{x}}{3\sqrt{x}}\)
b) \(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\) (*)
Thay (*) vào B , ta được : \(B=\dfrac{2-\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{-\sqrt{3}+1}{3\sqrt{3}+3}\)
Bạn santa làm sai r nha!
a, ĐKXĐ: x \(\ge\) 0; x \(\ne\) 4; x \(\ne\) 0
B = \(\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right)\)
B = \(\left(\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right)\)
B = \(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
B = \(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3}\)
B = \(\dfrac{\left(2-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{3\sqrt{x}\left(\sqrt{x}+1\right)}\)
B = \(\dfrac{2-\sqrt{x}}{3\sqrt{x}}\) (Đoạn này bạn kia viết sai đề mà vẫn đúng kết quả được?)
Vậy ...
b, Ta có: x = 4 + 2\(\sqrt{3}\) = (\(\sqrt{3}\) + 1)2 (TMĐK)
\(\Rightarrow\) \(\sqrt{x}\) = \(\sqrt{3}+1\) (1)
Thay (1) vào B ta được:
B = \(\dfrac{2-\sqrt{3}-1}{3\left(\sqrt{3}-1\right)}\) = \(\dfrac{1-\sqrt{3}}{-3\left(1-\sqrt{3}\right)}\) = \(\dfrac{-1}{3}\)
Vậy ...
Chúc bn học tốt!
mình làm lại nhé :
đkxđ : \(\left\{{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)
\(\Leftrightarrow B=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow B=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{3}\)
\(\Leftrightarrow B=\dfrac{2-\sqrt{x}}{3\sqrt{x}}\)
câu b làm như kia là oke rồi nhé <3
cho \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\) hãy tính giá trị của biểu thức sau:
\(A=f\left(\dfrac{1}{2012}\right)+f\left(\dfrac{2}{2012}\right)+...+f\left(\dfrac{2010}{2012}\right)+f\left(\dfrac{2011}{2012}\right)\)
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