Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x+2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=19\)
Chúc bạn học tốt!!!
a, \(\dfrac{x+1}{5}+\dfrac{x+3}{4}=\dfrac{x+5}{3}+\dfrac{x+7}{2}\)
\(\Rightarrow\dfrac{x+1}{5}+2+\dfrac{x+3}{4}+2=\dfrac{x+5}{3}+2+\dfrac{x+7}{2}+2\)
\(\Rightarrow\dfrac{x+11}{5}+\dfrac{x+11}{4}-\dfrac{x+11}{3}-\dfrac{x+11}{2}=0\)
\(\Rightarrow\left(x+11\right)\left(\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)
\(\Rightarrow x+11=0\Rightarrow x=-11\)
Vậy x = -11
b, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{15}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=15\)
Vậy x = 15
1) a) Ta có: \(\frac{x}{-15}=\frac{-60}{x}\) \(\Rightarrow x^2=\left(-15\right).\left(-60\right)=900\)
\(\Rightarrow x=30\)
b) \(\frac{-2}{x}=\frac{-x}{\frac{8}{25}}\) \(\Rightarrow x.\left(-x\right)=\left(-2\right).\frac{8}{25}\)
\(\Rightarrow x.\left(-x\right)=\frac{-16}{25}\)
\(\Rightarrow x.\left(-x\right)=\left(\frac{-4}{5}\right).\frac{4}{5}\)
Vậy \(x=\frac{4}{5}\)
2) a) \(3,8: \left(2x\right)=\frac{1}{4}:2\frac{2}{3}\)
\(\Rightarrow3,8: \left(2x\right)=\frac{3}{32}\)
\(\Rightarrow2x=\frac{3}{32}:3,8=\frac{15}{608}\)
\(x=\frac{15}{608}:2=\frac{15}{1216}\)
Vậy \(x=\frac{15}{1216}\)
b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)
\(\Rightarrow\left(0,25x\right):3=\frac{20}{3}\)
\(\Rightarrow0,25x=\frac{20}{3}.3=20\)
\(\Rightarrow x=20:0,25=80\)
Vậy x = 80
c) \(0,01:2,5=\left(0,75x\right):0,75\)
\(\Rightarrow\frac{1}{250}=\left(0,75x\right):0,75\)
\(\Leftrightarrow0,75x=\frac{1}{250}.0,75=\frac{3}{1000}\)
\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)
Vậy \(x=\frac{1}{250}\)
d) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow0,1x=\frac{5}{3}.\frac{2}{3}=\frac{10}{9}\)
\(\Rightarrow x=\frac{10}{9}:0,1=\frac{100}{9}\)
Vậy \(x=\frac{100}{9}\)
a) \(\frac{x}{-15}=\frac{-60}{x}\Leftrightarrow x.x=-15.\left(-60\right)\Leftrightarrow x^2=900\Leftrightarrow x^2=\orbr{\begin{cases}30^2\\\left(-30\right)^2\end{cases}}\Leftrightarrow x=\orbr{\begin{cases}30\\-30\end{cases}}\)
a) \(\left|x-\dfrac{5}{3}\right|< \dfrac{1}{3}\)
\(\Rightarrow\dfrac{-1}{3}< x-\dfrac{5}{3}< \dfrac{1}{3}\)
\(\Rightarrow\dfrac{-1}{3}+\dfrac{5}{3}< x-\dfrac{5}{3}+\dfrac{5}{3}< \dfrac{1}{3}+\dfrac{5}{3}\)
\(\Rightarrow\dfrac{4}{3}< x< 2\)
b) \(\left|x+\dfrac{11}{2}\right|>\left|-5,5\right|=5,5\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{11}{2}< 5,5\\x+\dfrac{11}{2}>5,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 5,5-\dfrac{11}{2}=0\\x>5,5-\dfrac{11}{2}=0\end{matrix}\right.\)
=> Với x khác 0 thì thõa mãn đề bài
c) \(\dfrac{2}{5}< \left|x-\dfrac{7}{5}\right|< \dfrac{3}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2}{5}< x-\dfrac{7}{5}< \dfrac{3}{5}\\-\dfrac{2}{5}< x-\dfrac{7}{5}< -\dfrac{3}{5}\end{matrix}\right.\)
Ta thấy trường hợp 2 là trường hợp không thể xảy ra
=> Loại
Vậy \(\dfrac{2}{5}< x-\dfrac{7}{5}< \dfrac{3}{5}\)
\(\Rightarrow\dfrac{2}{5}+\dfrac{7}{5}< x< \dfrac{3}{5}+\dfrac{7}{5}\)
\(\Rightarrow\dfrac{9}{5}< x< 2\) (nhận)
p/s : làm đại nha , ko bik đúng sai
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
~ Học tốt ~
Bài 1:
1) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)
\(=3^2.\left(\dfrac{1}{3}\right)^5.\left(3^4\right)^2.\dfrac{1}{3^3}\)
\(=3^2.\dfrac{1}{3^5}.3^8.\dfrac{1}{3^3}\)
\(=3^2=9\)
2) \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)
\(=\left(2^2.2^5\right):[2^3.\left(\dfrac{1}{2}\right)^4]\)
\(=2^7:2^3:\dfrac{1}{2^4}\)
\(=2^4.2^4=256\)
3)\(\left(2^{-1}+3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}.1:2^3\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2^4}\)
\(=\dfrac{43}{48}\)
4)\(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=-3-1+\dfrac{1}{4}.\dfrac{1}{2}\)
\(=-3-1+\dfrac{1}{8}\)
\(=-4+\dfrac{1}{8}\\ \)
\(=-\dfrac{31}{8}\)
5)\([\left(0,1\right)^2]^0+[\left(\dfrac{1}{7}\right)^{-1}]^2.\dfrac{1}{49}.[\left(2^2\right)^3:2^5]\\ =1+7^2.\dfrac{1}{7^2}.2^6:2^5\\ =1+1.2\\ =3\)
Chúc bạn học tốt 
a: Gọi số nguyên cần tìm là x
Theo đề, ta có: \(\dfrac{1}{3}+\left(\dfrac{2}{4}-1\dfrac{2}{5}\right)< x< 2\dfrac{1}{7}+\left(\dfrac{-2}{5}-\dfrac{1}{4}\right)\)
\(\Leftrightarrow\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{7}{5}< x< \dfrac{15}{7}-\dfrac{2}{5}-\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{20}{60}+\dfrac{30}{60}-\dfrac{84}{60}< x< \dfrac{15\cdot20-2\cdot28-35}{140}\)
\(\Leftrightarrow-\dfrac{34}{60}< x< \dfrac{209}{140}\)
mà x là số nguyên
nên \(x\in\left\{0;1\right\}\)
b: Gọi số nguyên cần tìm là x
Theo đề, ta có: \(\dfrac{7}{3}+\dfrac{3}{4}-\dfrac{1}{5}>x>\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{7\cdot20+3\cdot15-12}{60}>x>\dfrac{56-21+2\cdot12}{84}\)
\(\Leftrightarrow\dfrac{173}{60}>x>\dfrac{59}{84}\)
mà x là số nguên
nên \(x\in\left\{2;1\right\}\)
1)\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2017}{2018}\)
\(B=\dfrac{1}{2018}\)
2)a)\(x^2-2x-15=0\)
\(\Leftrightarrow x^2-2x+1-16=0\)
\(\Leftrightarrow\left(x-1\right)^2-16=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
3)\(\dfrac{a}{b}=\dfrac{d}{c}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a}{b}\cdot\dfrac{d}{c}=\dfrac{ad}{bc}\)
Lại có:\(\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a^2+d^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a^2+d^2}{b^2+c^2}=\dfrac{ad}{bc}\)
4)Ta có:\(g\left(x\right)=-x^{101}+x^{100}-x^{99}+...+x^2-x+1\)
\(g\left(x\right)=-x^{101}+\left(x^{100}-x^{99}+...+x^2-x+1\right)\)
\(g\left(x\right)=-x^{101}+f\left(x\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=f\left(x\right)+x^{101}-f\left(x\right)=x^{101}\)
Tại x=0 thì f(x)-g(x)=0
Tại x=1 thì f(x)-g(x)=1
làm bài 3 BĐT
theo bảng xét dấu
còn bài 1,2 ở trên là 1.1 và 1.2 đều trg bài 1.2
bài 1.2 (tức bài 2 ở trên )làm a,b,c,d
\còn bài 2( tức bài 2 ở trên) làm hết
1.
a) Tìm GTLN của A = \(\dfrac{6}{\left|x\right|-3}\)
Ta có : \(\left|x\right|\ge0\Rightarrow\left|x\right|-3\ge-3\)
\(\Rightarrow\dfrac{1}{\left|x\right|-3}\le\dfrac{1}{-3}\)
\(\Rightarrow\dfrac{6}{\left|x\right|-3}\le\dfrac{6}{-3}\)
\(\Rightarrow\dfrac{6}{\left|x\right|-3}\le-2\)
Hay A \(\le-2\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
Vậy MaxA = -2 \(\Leftrightarrow x=0\)
b) Tìm GTLN của \(E=\dfrac{6}{\left|x-2\right|+3}\)
Ta có : \(\left|x-2\right|\ge0\Rightarrow\left|x-2\right|+3\ge3\)
\(\Rightarrow\dfrac{1}{\left|x-2\right|+3}\le\dfrac{1}{3}\)
\(\Rightarrow\dfrac{6}{\left|x-2\right|+3}\le\dfrac{6}{3}=2\)
Hay \(E\le2\)
Dấu "=" xảy ra \(\Leftrightarrow x-2=0\Rightarrow x=2\)
Vậy MaxE = 2 <=> X =2
bài 3 :
Ta có : \(\left|a-b\right|=\left|\left(a-c\right)+\left(c-b\right)\right|\le\left|a-c\right|+\left|c-b\right|< 3+2=5\)
=> đpcm