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1,
Ta có: \(x^2\ge0;\left|y-13\right|\ge0\)
\(\Rightarrow x^2+\left|y-13\right|\ge0\)
\(\Rightarrow x^2+\left|y-13\right|+14\ge14\)
\(\Rightarrow\frac{1}{x^2+\left|y-13\right|+14}\le\frac{1}{14}\)
\(\Rightarrow P=\frac{12}{x^2+\left|y-13\right|+14}\le\frac{12}{14}=\frac{6}{7}\)
Dấu "=" xảy ra khi x = 0, y = 13
Vậy Pmin = 6/7 khi x = 0, y = 13
2, \(P=\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=1+\frac{7}{n-5}\)
Để P có GTLN thì\(\frac{7}{n-5}\) có GTLN => n - 5 có GTNN và n - 5 > 0 => n = 6
3,
Ta có: \(10\le n\le99\)
\(\Rightarrow20\le2n\le198\)
\(\Rightarrow2n\in\left\{36;64;100;144;196\right\}\)
\(\Rightarrow n\in\left\{18;32;50;72;98\right\}\)
\(\Rightarrow n+4\in\left\{22;36;50;72;98\right\}\)
Ta thấy chỉ có 36 là số chính phương
Vậy n = 32
4,
ÁP dụng TCDTSBN ta có:
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{a+c-b}{b}=\frac{a+b-c+b+c-a+a+c-b}{c+a+b}=\frac{a+b+c}{a+b+c}=1\) (vì a+b+c khác 0)
\(\Rightarrow\hept{\begin{cases}\frac{a+b-c}{c}=1\\\frac{b+c-a}{a}=1\\\frac{a+c-b}{b}=1\end{cases}\Rightarrow\hept{\begin{cases}a+b-c=c\\b+c-a=a\\a+c-b=b\end{cases}\Rightarrow}\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}}\)
\(\Rightarrow B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}\cdot\frac{a+c}{c}\cdot\frac{b+c}{b}=\frac{2c}{a}\cdot\frac{2b}{c}\cdot\frac{2a}{b}=\frac{8abc}{abc}=8\)
Vậy B = 8
Bài 1 và Bài 2 dễ, bn có thể tự làm được!
Bài 3:
a) ta có: 1020 = (102)10 = 10010
=> 10010>910
=> 1020>910
b) ta có: (-5)30 = 530 =( 53)10 = 12510 ( vì là lũy thừa bậc chẵn)
(-3)50 = 350 = (35)10= 24310
=> 12510 < 24310
=> (-5)30 < (-3)50
c) ta có: 648 = (26)8= 248
1612 = ( 24)12 = 248
=> 648 = 1612
d) ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
a: =>|x-1/4|=3/4
=>x-1/4=3/4 hoặc x-1/4=-3/4
=>x=1 hoặc x=-1/2
b: \(\left|x+\dfrac{1}{2}\right|=\dfrac{1}{2}-\dfrac{9}{4}=\dfrac{2-9}{4}=-\dfrac{7}{4}\)(vô lý)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+5=1-x\\2x+5=x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\x=-6\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{4}{3};-6\right\}\)
e: =>|3/2-x|=0
=>3/2-x=0
hay x=3/2
a,\(\left(x-\frac{2}{3}\right),\left(x+\frac{1}{1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{2}{3}\\x+\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{-1}{4}\end{matrix}\right.\)
b,\(\left(x-\frac{2}{3}\right)\left(2x-\frac{3}{4}\right)=\left(3x+\frac{1}{2}\right)\left(x+\frac{2}{3}\right)\)
\(\Leftrightarrow2x^2-\frac{3}{4}x-\frac{4}{3}x+\frac{1}{2}=3x^2+2x+\frac{1}{2}x+\frac{1}{3}\)
\(\Leftrightarrow2x^2-\frac{25}{12}x+\frac{1}{2}=3x^2+\frac{5}{2}x+\frac{1}{3}\)
\(\Leftrightarrow24x^2-25x+6=36x^2+30x+4\)
\(\Leftrightarrow24x^2-25x+6-36x^2-30x-4=0\)
\(\Leftrightarrow-12x^2-55x+2=0\)
\(\Leftrightarrow12x^2+55x-2=0\)
\(\Leftrightarrow x=\frac{-55\pm\sqrt{55^2-4.12\left(-2\right)}}{2.12}\)
\(\Leftrightarrow\frac{-55\pm\sqrt{3025+96}}{24}\)
\(\Leftrightarrow\frac{-55\pm\sqrt{3121}}{24}\)
\(\Leftrightarrow\frac{-55+\sqrt{3121}}{24}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{-55+\sqrt{3121}}{24}\\\frac{-55-\sqrt{3121}}{24}\end{matrix}\right.\)
Bài 2:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)
Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)
\(\Rightarrow4x+12=6x\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
Vậy x = 6
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)
\(=\frac{14-5}{8}=\frac{9}{8}\)
+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)
+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)
+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)
Vậy ...
c) \(5^x+5^{x+1}+5^{x+2}=3875\)
\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)
\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)
\(\Rightarrow5^x.31=3875\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy x = 3
a) => 4x + 2/3 = 0 hoặc 2/3x - 1 =0
4x= -2/3 hoặc 2/3x= 1
x = -2/3 . 1/4 hoặc x = 1.3/2
x = -1/6 hoặc x = 3/2
b) x+2 / x -1 = 5/2
=> 2(x+2) = 5(x-1)
2x + 4 = 5x - 5
5x - 2x= 4+5
3x = 9
=> x= 3
a) (4x+\(\frac{2}{3}\)) . ( \(\frac{2}{3}\)x-1)=0
\(\Rightarrow\)\(\orbr{\begin{cases}4x+\frac{2}{3}=0\\\frac{2}{3}x-1=0\end{cases}}\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=\\x=\end{cases}}\)........
Tới đây bn tự giải nha
a) \(3,6-\left|x-0,4\right|=0\)
\(\Leftrightarrow\left|x-0,4\right|=3,6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)
Vậy \(x\in\left\{4;-3,2\right\}\)
b) Ta có:
\(\frac{x}{2}=y=\frac{z}{3}=\frac{2y}{2}=\frac{x-2y+z}{2-2+3}=\frac{210}{3}=70\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=70\\y=70\\\frac{z}{3}=70\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=140\\y=70\\z=210\end{matrix}\right.\)
Vậy \(x=140\); \(y=70\); \(z=210\)
c)\(\left|x+0,25\right|-4=\frac{1}{4}\)
\(\Leftrightarrow\left|x+\frac{1}{4}\right|=\frac{17}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{17}{4}\\x+\frac{1}{4}=\frac{-17}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{-9}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{4;\frac{-9}{2}\right\}\)
d) \(x:\left(0,25\right)^4=\left(0,5\right)^2\)
\(\Leftrightarrow x=\left(0,25\right)^4.\left(0,5\right)^2\)
\(\Leftrightarrow x=\left(0,5\right)^8.\left(0,5\right)^2\)
\(\Leftrightarrow x=\left(0,5\right)^{10}=\left(\frac{1}{2}\right)^{10}=\frac{1}{2^{10}}=\frac{1}{1024}\)
Vậy \(x=\frac{1}{1024}\)
e) \(3^{x-1}+5.3^{x-1}=162\)
\(\Leftrightarrow6.3^{x-1}=162\)
\(\Leftrightarrow3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
f) \(\frac{x}{-25}=\frac{2}{5}\)
\(\Leftrightarrow x=\left(-25\right).\frac{2}{5}=-10\)
Vậy \(x=-10\)
g) \(\left|x+\frac{3}{4}\right|-\frac{3}{4}=\sqrt{\frac{1}{9}}\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|-\frac{3}{4}=\frac{1}{3}\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{13}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{13}{12}\\x+\frac{3}{4}=-\frac{13}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-\frac{11}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{3};-\frac{11}{6}\right\}\)
a) \(3,6-\left|x-0,4\right|=0\)
\(\Rightarrow\left|x-0,4\right|=3,6-0\)
\(\Rightarrow\left|x-0,4\right|=3,6.\)
\(\Rightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,6+0,4\\x=\left(-3,6\right)+0,4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)
Vậy \(x\in\left\{4;-3,2\right\}.\)
c) \(\left|x+0,25\right|-4=\frac{1}{4}\)
\(\Rightarrow\left|x+\frac{1}{4}\right|=\frac{1}{4}+4\)
\(\Rightarrow\left|x+\frac{1}{4}\right|=\frac{17}{4}.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{17}{4}\\x+\frac{1}{4}=-\frac{17}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{17}{4}-\frac{1}{4}\\x=\left(-\frac{17}{4}\right)-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{9}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{4;-\frac{9}{2}\right\}.\)
d) \(x:\left(0,25\right)^4=\left(0,5\right)^2\)
\(\Rightarrow x:\left(0,25\right)^4=0,25\)
\(\Rightarrow x=\left(0,25\right).\left(0,25\right)^4\)
\(\Rightarrow x=\left(0,25\right)^5\)
\(\Rightarrow x=\frac{1}{1024}\)
Vậy \(x=\frac{1}{1024}.\)
Chúc bạn học tốt!
a) Quy đồng lên đi.
b) \(\frac{x+2}{0.5}=\frac{2x+1}{2}\Leftrightarrow\frac{x+2}{\left(\frac{1}{2}\right)}=\frac{2x+1}{2}\)
\(\Leftrightarrow2x+4=\frac{2x+1}{2}\Leftrightarrow4x+8=2x+1\)
\(\Leftrightarrow x=-\frac{7}{2}\)
c) \(\Leftrightarrow\left|x+\frac{1}{5}\right|=6\). VỚi x >= -1/5 thì:
\(x+\frac{1}{5}=6\Leftrightarrow x=\frac{29}{5}\left(TM\right)\)
Với x < -1/5 thì \(-x-\frac{1}{5}=6\Leftrightarrow x=-\frac{31}{5}\left(TM\right)\)
d) TƯơng tự ý a, quy đồng lên thôi (mẫu chung là 24 thì phải)
c) \(\left|x+\frac{1}{5}\right|-4=2\)
=> \(\left|x+\frac{1}{5}\right|=2+4\)
=> \(\left|x+\frac{1}{5}\right|=6\)
=> \(\left\{{}\begin{matrix}x+\frac{1}{5}=6\\x+\frac{1}{5}=-6\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=6-\frac{1}{5}\\x=\left(-6\right)-\frac{1}{5}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{29}{5}\\x=-\frac{31}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{29}{5};-\frac{31}{5}\right\}\).
Mình chỉ làm câu c) thôi nhé.
Chúc bạn học tốt!