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bài/2.1/bài dùng chặn
AT
8 tháng 6 2021
2.1) \(A=\dfrac{\sqrt{x}+7}{\sqrt{x}+2}=1+\dfrac{5}{\sqrt{x}+2}>1\)
Ta có \(\sqrt{x}+2\ge2\Rightarrow\dfrac{5}{\sqrt{x}+2}\le\dfrac{5}{2}< 3\Rightarrow A< 4\)
mà \(A\in Z\Rightarrow\left[{}\begin{matrix}A=2\\A=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{5}{\sqrt{x}+2}=1\\\dfrac{5}{\sqrt{x}+2}=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\sqrt{x}+2=5\\\sqrt{x}+2=\dfrac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{1}{4}\end{matrix}\right.\)
AT
29 tháng 5 2021
1.\(A=\left(\sqrt{3}+1\right)\sqrt{\dfrac{14-6\sqrt{3}}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\dfrac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\dfrac{44\left(2-\sqrt{3}\right)}{22}}=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)
AT
29 tháng 5 2021
2.1.a) \(x^2=\left(x-1\right)\left(3x-2\right)\Leftrightarrow x^2=3x^2-5x+2\Leftrightarrow2x^2-5x+2=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
b) \(9x^4+5x^2-4=0\Leftrightarrow9x^4+9x^2-4x^2-4=0\)
\(\Leftrightarrow9x^2\left(x^2+1\right)-4\left(x^2+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(9x^2-4\right)=0\)
mà \(x^2+1>0\Rightarrow9x^2=4\Rightarrow x^2=\dfrac{4}{9}\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2) Gọi số xe lúc đầu của đội là a(xe) \(\left(a\in N,a>0\right)\)
Theo đề,ta có: \(\left(a-2\right)\left(\dfrac{120}{a}+3\right)=120\Leftrightarrow120+3a-\dfrac{240}{a}-6=120\)
\(\Leftrightarrow\dfrac{3a^2-6a-240}{a}=0\Rightarrow3a^2-6a-240=0\Rightarrow a^2-2a-80=0\)
\(\Leftrightarrow\left(a+8\right)\left(a-10\right)=0\) mà \(a>0\Rightarrow a=10\)
T
1
8 tháng 1 2016
\(\frac{1}{100}-\frac{1}{100\cdot99}-\frac{1}{99.98}-...-\frac{1}{2\cdot1}\)
\(=\frac{1}{100}-\frac{1}{99}-\frac{1}{100}-\frac{1}{99}-\frac{1}{98}-...-1-\frac{1}{2}\)
\(=1\)
C
2
29 tháng 8 2023
1: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)
\(=3+2\sqrt{2}+\sqrt{5}-2=2\sqrt{2}+\sqrt{5}+1\)
2: \(\dfrac{1}{\sqrt{3}+\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}+\dfrac{2\left(1+\sqrt{7}\right)}{-6}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}-\dfrac{1+\sqrt{7}}{3}\)
\(=\dfrac{3\left(\sqrt{7}-\sqrt{3}\right)-4\left(\sqrt{7}+1\right)}{12}=\dfrac{-\sqrt{7}-3\sqrt{3}-4}{12}\)
3:
\(=\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{2-\sqrt{a}}=-\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}=-\sqrt{a}\)
4:
\(=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{xy}\)
H9
HT.Phong (9A5)
CTVHS
29 tháng 8 2023
1) \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=\dfrac{3+2\sqrt{2}}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}+\dfrac{\sqrt{5}-2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)
\(=\dfrac{3+2\sqrt{2}}{3^2-\left(2\sqrt{2}\right)^2}+\dfrac{\sqrt{5}-2}{\left(\sqrt{5}\right)^2-2^2}\)
\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)
\(=3+2\sqrt{2}+\sqrt{5}-2\)
\(=2\sqrt{2}+\sqrt{5}+1\)
2) \(\dfrac{1}{\sqrt{3}-\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)
\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{3}-\sqrt{7}\right)}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{\left(1-\sqrt{7}\right)\left(1+\sqrt{7}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}\right)^2-\left(\sqrt{7}\right)^2}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{1^2-\left(\sqrt{7}\right)^2}\)
\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{2\cdot\left(1+\sqrt{7}\right)}{6}\)
\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{1+\sqrt{7}}{3}\)
\(=\dfrac{-3\sqrt{3}-3\sqrt{7}}{12}-\dfrac{4+4\sqrt{7}}{12}\)
\(=\dfrac{-3\sqrt{3}-3\sqrt{7}-4-4\sqrt{7}}{12}\)
\(=\dfrac{-3\sqrt{3}-7\sqrt{7}-4}{12}\)
3) \(\dfrac{a-2\sqrt{a}}{2-\sqrt{a}}\)
\(=-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\)
\(=-\dfrac{\sqrt{a}\cdot\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\)
\(=-\sqrt{a}\)
4) \(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{xy}+\sqrt{y}\cdot\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{xy}\cdot\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{xy}\)
H
0
29 tháng 6 2017
Ta có:
\(\frac{2n+1}{\left[n\left(n+1\right)\right]^2}=\frac{n+n+1}{n^2\left(n+1\right)^2}=\frac{1}{n\left(n+1\right)^2}+\frac{1}{n^2\left(n+1\right)}\)
\(=\frac{1}{n\left(n+1\right)}.\left(\frac{1}{n}+\frac{1}{n+1}\right)=\left(\frac{1}{n}-\frac{1}{n+1}\right).\left(\frac{1}{n}+\frac{1}{n+1}\right)\)
\(=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
Áp dụng vào bài toán ta được
\(A=\frac{2.1+1}{\left[1\left(1+1\right)\right]^2}+\frac{2.2+1}{\left[2\left(2+1\right)\right]^2}+...+\frac{2.99+1}{\left[99\left(99+1\right)\right]^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}\)
\(=1-\frac{1}{100^2}=\frac{9999}{10000}\)
17 tháng 12 2021
1.
\(a,\Leftrightarrow2m-1+m-2=6\Leftrightarrow3m=9\Leftrightarrow m=3\\ b,2x+3y-5=0\Leftrightarrow3y=-2x+5\Leftrightarrow y=-\dfrac{2}{3}x+\dfrac{5}{3}\)
Để \(\left(d\right)\text{//}y=-\dfrac{2}{3}x+\dfrac{5}{3}\Leftrightarrow\left\{{}\begin{matrix}2m-1=-\dfrac{2}{3}\\m-2\ne\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=\dfrac{1}{6}\\m\ne\dfrac{11}{3}\end{matrix}\right.\Leftrightarrow m=\dfrac{1}{6}\)
\(c,x+2y+1=0\Leftrightarrow2y=-x-1\Leftrightarrow y=-\dfrac{1}{2}x-\dfrac{1}{2}\\ \left(d\right)\bot y=-\dfrac{1}{2}x-\dfrac{1}{2}\Leftrightarrow\left(-\dfrac{1}{2}\right)\left(2m-1\right)=-1\\ \Leftrightarrow\dfrac{1}{2}\left(2m-1\right)=1\Leftrightarrow m-\dfrac{1}{2}=1\Leftrightarrow m=\dfrac{3}{2}\)
2.
Gọi điểm cố định đó là \(A\left(x_0;y_0\right)\)
\(\Leftrightarrow y_0=\left(2m-1\right)x_0+m-2\\ \Leftrightarrow2mx_0+m-x_0-2-y_0=0\\ \Leftrightarrow m\left(2x_0+1\right)-\left(x_0+y_0+2\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x_0=-1\\x_0+y_0+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_0=-\dfrac{1}{2}\\y_0=-\dfrac{3}{2}\end{matrix}\right.\)
\(x^2-2\left(m-1\right)x+2m-4=0\)
\(\Delta'=\left[-\left(m-1\right)\right]^2-\left(2m-4\right)\)
\(\Delta'=m^2-2m+1-2m+4\)
\(\Delta'=m^2-4m+5\)
\(\Delta'=\left(m-2\right)^2+1>0\)(với mọi m)
Vậy pt luôn có 2 nghiệm x1 ,x2 với mọi m
theo đinh lí Vi-ét ta có
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)=2m-2\\x_1x_2=2m-4\end{matrix}\right.\)
ta có \(x_1^2+x_2^2=3\)
⇔\(\left(x_1+x_2\right)^2-2x_1x_2=3\)
⇔\(\left(2m-2\right)^2-2\left(2m-4\right)=3\)
⇔\(4m^2-8m+4-4m+8=3\)
⇔\(4m^2-12m+9=0\)
\(\Delta'=\left(-6\right)^2-4\cdot9=0\)
⇒Pt có nghiệm kép
\(m_1=m_2=\dfrac{6}{4}=\dfrac{3}{2}\)
vậy m =\(\dfrac{3}{2}\)
\(x^2-2\left(m-1\right)x+2m-4=0\)(1)
\(\Delta'=\left[-\left(m-1\right)\right]^2-\left(2m-4\right)=m^2-2m+1-2m+4\)
\(=m^2-4m+5=\left(m-2\right)^2+1>0\left(\forall m\right)\)
\(=>\Delta'>0\) nên pt (1) luôn có 2 nghiệm phân biệt \(\forall m\\ \)
theo vi ét=>\(\left\{{}\begin{matrix}x1+x2=2\left(m-1\right)=2m-2\\x1x2=2m-4\end{matrix}\right.\)
có \(x1^2+x2^2=3< =>\left(x1+x2\right)^2-2x1x2-3=0\)
\(< =>\left(2m-2\right)^2-2\left(2m-4\right)-3=0\)
\(< =>4m^2-8m+4-4m+5=0\)
\(< =>4m^2-12m+9=0\)(2)
\(\Delta1=\left(-12\right)^2-4.9.4=0\)
=>pt (2) có nghiệm kép m1=m2=\(\dfrac{12}{2.4}=\dfrac{12}{8}=1,5\)
vậy m=1,5....