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MT
1
DT
6 tháng 3 2022
a) A = \(\sum\limits^{50}_1\left(2x\right)-\sum\limits^{50}_1\left(2x-1\right)\) = 5050
b) B = \(\sum\limits^{2010}_1x^3\) = 4084663313000
27 tháng 9 2021
b: \(\sqrt{8-2\sqrt{15}}-\sqrt{5}\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{5}\)
\(=-\sqrt{3}\)
c: \(\sqrt{11-6\sqrt{2}}=3-\sqrt{2}\)
d: \(\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2022
\(\Delta'=4-\left(m+1\right)\ge0\Rightarrow m\le3\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=m+1\end{matrix}\right.\)
\(x_1^2+x_2^2=5\left(x_1+x_2\right)\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=5\left(x_1+x_2\right)\)
\(\Leftrightarrow16-2\left(m+1\right)=20\)
\(\Leftrightarrow m=-3\) (thỏa mãn)
AM
7 tháng 2 2022
a. Ta có: \(x^2-4x+m+1=0\)
Thay m=2 ta được: \(x^2-4x+2+1=0\Leftrightarrow x^2-4x+3=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
b. Để phương trình có 2 nghiệm phân biệt thì \(\Delta=\left(-4\right)^2-4.1.\left(m+1\right)>0\)
\(\Leftrightarrow16-4\left(m+1\right)>0\Leftrightarrow16>4\left(m+1\right)\Leftrightarrow4>m+1\Leftrightarrow m< 3\)
Áp dụng định lí Vi-et ta có: \(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=m+1\end{matrix}\right.\)
Theo đề ta có: \(x_1^2+x_2^2=5\left(x_1+x_2\right)\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=5\left(x_1+x_2\right)\)
\(\Leftrightarrow\left(4\right)^2-2\left(m+1\right)=5.4\)
\(\Leftrightarrow16-2m-2=20\Leftrightarrow m=-3\) (TM)
27 tháng 2 2022
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{3}y=\dfrac{7}{3}\\x-\dfrac{1}{2}y=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{6}y=\dfrac{5}{2}\\x+\dfrac{1}{3}y=\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{4}{3}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
27 tháng 2 2022
Lời giải:
Lấy PT(1) trừ PT(2) theo vế:
$\frac{y}{3}+\frac{y}{2}=\frac{7}{3}+\frac{1}{6}$
$\Leftrightarrow \frac{5}{6}y=\frac{5}{2}$
$\Leftrightarrow y=3$
$x=\frac{7}{3}-\frac{y}{3}=\frac{7}{3}-1=\frac{4}{3}$
22 tháng 6 2018
Ta có: \(\hept{\begin{cases}\left(\frac{1}{x}+y\right)+\left(\frac{1}{x}-y\right)=\frac{5}{8}\\\left(\frac{1}{x}+y\right)-\left(\frac{1}{x}-y\right)=-\frac{3}{8}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2}{x}=\frac{5}{8}\\2y=-\frac{3}{8}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{16}{5}\\y=-\frac{3}{16}\end{cases}}}\)
ND
Nguyễn Đức Trí
VIP
21 tháng 9 2023
a) \(\dfrac{1}{\sqrt[]{x}-1}+\dfrac{1}{1+\sqrt[]{x}}+1\left(x\ge0;x\ne1\right)\)
\(=\dfrac{\sqrt[]{x}+1+\sqrt[]{x}-1+x-1}{\left(\sqrt[]{x}-1\right)\left(\sqrt[]{x}+1\right)}\)
\(=\dfrac{x+2\sqrt[]{x}-1}{x-1}\)
\(=\dfrac{x-1+2\sqrt[]{x}}{x-1}\)
\(=1+\dfrac{2\sqrt[]{x}}{x-1}\)
b) \(\dfrac{1}{\sqrt[]{x}+2}-\dfrac{2}{\sqrt[]{x}-2}-\dfrac{4}{4-x}\left(x\ge0;x\ne4\right)\)
\(=\dfrac{\sqrt[]{x}-2-2\left(\sqrt[]{x}+2\right)+4}{\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}-2\right)}\)
\(=\dfrac{\sqrt[]{x}-2-2\sqrt[]{x}-4+4}{\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}-2\right)}\)
\(=\dfrac{-\sqrt[]{x}-2}{\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}-2\right)}\)
\(=\dfrac{-\left(\sqrt[]{x}+2\right)}{\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}-2\right)}\)
\(=\dfrac{-1}{\sqrt[]{x}-2}\)
9 tháng 5 2023
a: Thay x=2 vào (P),ta được:
y=2^2/2=2
2: Thay x=2 và y=2 vào (d), ta được:
m-1+2=2
=>m-1=0
=>m=1
3)
a) Ta có:
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{ac}+\dfrac{2}{bc}\\ =\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2a+2b+2c}{abc}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2\left(a+b+c\right)}{abc}\\ =\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
\(=>\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\)
b)
\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2024^2}+\dfrac{1}{2025^2}}\left(1\right)\\ =\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{\left(-3\right)^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{\left(-4\right)^2}}+...+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2024^2}+\dfrac{1}{\left(-2025\right)^2}}\)
Theo câu a \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\) khi \(a+b+c=0\)
Mà: \(\left\{{}\begin{matrix}1+2+\left(-3\right)=0\\1+3+\left(-4\right)=0\\....\\1+2024+\left(-2025\right)=0\end{matrix}\right.\)
=> \(\left(1\right)=\left|1+\dfrac{1}{2}+\dfrac{1}{-3}\right|+\left|1+\dfrac{1}{3}+\dfrac{1}{-4}\right|+...+\left|1+\dfrac{1}{2024}+\dfrac{1}{-2025}\right|\)
Mà: \(\left\{{}\begin{matrix}1+\dfrac{1}{2}\cdot\dfrac{1}{-3}>0\\1+\dfrac{1}{3}+\dfrac{1}{-4}>0\\...\\1+\dfrac{1}{2024}+\dfrac{1}{-2025}>0\end{matrix}\right.\)
=> \(\left(1\right)=1+\dfrac{1}{2}+\dfrac{1}{-3}+1+\dfrac{1}{3}+\dfrac{1}{-4}+...+1+\dfrac{1}{2024}+\dfrac{1}{-2025}\\ =\left(1+1+...+1\right)+\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{2024}-\dfrac{1}{2024}\right)-\dfrac{1}{2025}\\ =2023+\dfrac{1}{2}-\dfrac{1}{2025}\)
4)
\(\left(12-6\sqrt{3}\right)\sqrt{\dfrac{3}{14-8\sqrt{3}}}-3\sqrt{2\left(1-\sqrt{-2\sqrt{3}+4}\right)+2\sqrt{4+2\sqrt{3}}}\\ =\left(12-6\sqrt{3}\right)\sqrt{\dfrac{3}{\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2}\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}}-3\sqrt{2\left(1-\sqrt{4-2\sqrt{3}}\right)+2\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}}\\ =\left(12-6\sqrt{3}\right)\sqrt{\dfrac{3}{\left(2\sqrt{2}-\sqrt{6}\right)^2}}-3\sqrt{2\left(1-\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}\right)+2\sqrt{\left(\sqrt{3}+1\right)^2}}\\ =\left(12-6\sqrt{3}\right)\dfrac{\sqrt{3}}{2\sqrt{2}-\sqrt{6}}-3\sqrt{2\left(1-\sqrt{\left(\sqrt{3}-1\right)^2}\right)+2\left(\sqrt{3}+1\right)}\\ =\left(12-6\sqrt{3}\right)\dfrac{\sqrt{3}}{2\sqrt{2}-\sqrt{6}}-3\sqrt{2\left(1-\sqrt{3}+1\right)+2\sqrt{3}+2}\\ =\left(12-6\sqrt{3}\right)\dfrac{\sqrt{3}}{2\sqrt{2}-\sqrt{6}}-3\sqrt{2-2\sqrt{3}+2+2\sqrt{3}+2}\\ =3\sqrt{2}\cdot\left(2\sqrt{2}-\sqrt{6}\right)\cdot\dfrac{\sqrt{3}}{2\sqrt{2}-\sqrt{6}}-3\sqrt{6}\\ =3\sqrt{2}\cdot\sqrt{3}-3\sqrt{6}\\ =3\sqrt{6}-3\sqrt{6}\\ =0\)