giúp mik vs B1: a, x^2 : 16/11 = 1/4
b, / x-2 / – 3 = 0
c, 3 – / 1/3 – x / = 5/17
d, 1/3x + 1/5x = 2/3
B2: Cho A=1/2^2 + 1/3^2 + 1/4^2 + … + 1/100^2 Chứng minh rằng A < 3/4
Đáp án:
$a)
x=\pm \dfrac{2\sqrt{11}}{11}\\
b)
{\left[\begin{aligned}x=5\\x=-1\end{aligned}\right.}\\
c)
{\left[\begin{aligned} x=\dfrac{-121}{51}\\ x=\dfrac{155}{51}\end{aligned}\right.}\\
d)
x=\dfrac{5}{4}$
Giải thích các bước giải:
$a)
x^2:\dfrac{16}{11}=\dfrac{1}{4}\\
\Leftrightarrow x^2.\dfrac{11}{16}=\dfrac{1}{4}\\
\Leftrightarrow x^2=\dfrac{1}{4}.\dfrac{16}{11}=\dfrac{4}{11}\\
\Leftrightarrow x=\pm \dfrac{2\sqrt{11}}{11}\\
b)
\left | x-2 \right |-3=0\\
\Leftrightarrow \left | x-2 \right |=3\\
\Leftrightarrow {\left[\begin{aligned}x-2=3\\x-2=-3\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=5\\x=-1\end{aligned}\right.}\\
c)
3-\left | \dfrac{1}{3}-x \right |=\dfrac{5}{17}\\
\Leftrightarrow \left | \dfrac{1}{3}-x \right |=3-\dfrac{5}{17}\\
\Leftrightarrow \left | \dfrac{1}{3}-x \right |=\dfrac{51}{17}-\dfrac{5}{17}\\
\Leftrightarrow \left | \dfrac{1}{3}-x \right |=\dfrac{46}{17}\\
\Leftrightarrow {\left[\begin{aligned} \dfrac{1}{3}-x=\dfrac{46}{17}\\ \dfrac{1}{3}-x=\dfrac{-46}{17}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{1}{3}-\dfrac{46}{17}\\ x=\dfrac{1}{3}+\dfrac{46}{17}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{17}{51}-\dfrac{46.3}{51}\\ x=\dfrac{17}{51}+\dfrac{46.3}{51}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{-121}{51}\\ x=\dfrac{155}{51}\end{aligned}\right.}\\
d)
\dfrac{1}{3}x+\dfrac{1}{5}x=\dfrac{2}{3}\\
\Leftrightarrow \left ( \dfrac{1}{3}+\dfrac{1}{5} \right )x=\dfrac{2}{3}\\
\Leftrightarrow \left ( \dfrac{5}{15}+\dfrac{3}{15} \right )x=\dfrac{2}{3}\\
\Leftrightarrow \dfrac{8}{15}x=\dfrac{2}{3}\\
\Leftrightarrow x=\dfrac{2}{3}.\dfrac{15}{8}=\dfrac{5}{4}\\
2)
A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+…+\dfrac{1}{100^2}$
Ta có
$\dfrac{1}{2^2}=\dfrac{1}{4}\\
\dfrac{1}{3^2}<\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\\
\dfrac{1}{4^2}<\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}\\
….\\
\dfrac{1}{100^2}<\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\\
\Rightarrow A<\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{99}-\dfrac{1}{100}\\
\Rightarrow A<\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{100}\\
\Rightarrow A<\dfrac{3}{4}-\dfrac{1}{100}<\dfrac{3}{4} \left ( \dfrac{1}{100}>0 \right )\\
\Rightarrow A<\dfrac{3}{4}$