Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
x+7/2010+x+6/2011=x+5/2012+x+4/2013
((x+7/2010)-1)+((x+6/2011)-1)=(x+5/2012)-1)+(x+4/2013)-1)
x+2017/2010+x+2017/2011-x+2017/2012-x+2017/2013=0
x+2017(1/2010+1/2011-1/2012-1/2013)=0
x+2017=0(vì 1/2010+1/2011-1/2012-1/2013<0)
x=-2017
vậy.......
tk mk nha bn
a) \(\frac{2}{3a}-\frac{3}{a}=\frac{2}{3a}-\frac{9}{3a}=\frac{-7}{3a}=\frac{7}{15}\Leftrightarrow-3a=15\Leftrightarrow a=-5\)
b)\(2x^3-1=15\Leftrightarrow2x^3=16\Leftrightarrow x^3=8\Leftrightarrow x=2\)
\(\Rightarrow\frac{2+16}{9}=\frac{y-15}{16}=2\Leftrightarrow y-15=32\Leftrightarrow y=47\)
c) \(\left|x\right|=3\Rightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\) rồi xét 2 trường hợp để tính A nhé :)
Bài 1: ĐK của a: \(a\ne0\)
Quy đồng VT ta có: \(\frac{2a-9a}{3a^2}=\frac{7}{15}\)
\(\Leftrightarrow\frac{-7a}{3a^2}=\frac{7}{15}\)
\(\Leftrightarrow-7a.15=3a^2.7\)
\(\Leftrightarrow-105a=21a^2\)
\(\Leftrightarrow-105a-21a^2=0\)
\(\Leftrightarrow a\left(-105-21a\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}a=0\left(l\right)\\-105-21a=0\end{cases}\Leftrightarrow a=-5\left(n\right)}\)
Vậy:..
\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)
\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)
\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)
\(\Leftrightarrow x+116=0\Leftrightarrow x=-116\)
\(\frac{x-1}{117}+\frac{x-2}{118}+\frac{x-3}{119}=\frac{x-4}{120}+\frac{x-5}{121}+\frac{x-6}{122}\)
\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}+1=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)
\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)
\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)
Vì \(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\ne0\)
Nên x + 116 = 0
<=> x = -116
- Ta chứng minh bất đẳng thức phụ dưới đây: \(\frac{1}{\sqrt{x}\left(x+1\right)}=\frac{\sqrt{x}}{x\left(x+1\right)}=\sqrt{x}\left(\frac{1}{x}-\frac{1}{x+1}\right)=\sqrt{x}\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x+1}}\right)\)\(=\left(1+\frac{\sqrt{x}}{\sqrt{x+1}}\right)\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)< 2\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)\)
Áp dụng : \(\frac{1}{\sqrt{1}.2}< 2.\left(1-\frac{1}{\sqrt{2}}\right)\)
\(\frac{1}{\sqrt{2}.3}< 2.\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)\)
...................................
\(\frac{1}{\sqrt{2015}.2016}< 2.\left(\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)\)
Cộng các BĐT trên với nhau được : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}}< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)=2\left(1-\frac{1}{\sqrt{2016}}\right)< 2\left(1-\frac{1}{\sqrt{2025}}\right)=\frac{88}{45}\)
Từ đó suy ra đpcm
Cái ............... là gì vậy bn
Bài 1: Tìm x, y, z
\(\frac{x}{3}=\frac{y}{4}=>\frac{x}{3\times3}=\frac{y}{4\times3}=>\frac{x}{9}=\frac{y}{12}\)
\(\frac{y}{3}=\frac{z}{5}=>\frac{y}{3.4}=\frac{z}{5.4}=>\frac{y}{12}=\frac{z}{20}\)
=> \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)
- Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\) -> \(\frac{2x}{2\times9}=\frac{3y}{3\times12}=\frac{z}{20}\) -> \(\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}\)
-> \(\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)
\(\frac{x}{9}=3\rightarrow x=27\)
\(\frac{y}{12}=3\rightarrow y=36\)
\(\frac{z}{20}=3\rightarrow z=60\)
Vậy x = 27 ; y = 36 ; z = 60
Bài 2 : Tìm x, y:
5x = 2y và x.y = 40
Vì 5x = 2y => \(\frac{x}{2}=\frac{y}{5}\)
Cách 1:
\(\frac{x}{2}=\frac{y}{5}\) và x.y = 40
Đặt \(\frac{x}{2}=\frac{y}{5}\) = k
=> x = 2.k ; y = 5.k
x.y = 40 -> 2k = 5k = 40
-> 10 . \(k^2\) = 40
-> \(k^2\) = 4 -> k = 2 hoặc k = -2
k = 4 ta có : \(\frac{x}{2}=\frac{y}{5}=2->x=4;y=10\)
k = -4 ta có : \(\frac{x}{2}=\frac{y}{5}=-2->x=-4;y=-10\)
Cách 2:
\(\frac{x}{2}=\frac{y}{5}->\frac{x.x}{2}=\frac{x.y}{5}->\frac{x^2}{2}=\frac{40}{5}=\frac{x^2}{2}=8\)
=> \(x^2\) = 8 . 2 = 16 -> x = 4 hoặc -4
x = 4 -> 4.y = 40 => y = 10
x = -4 -> (-4).y = 40 => y = -10
Vậy x = 4 hoặc -4
y = 10 hoặc -10
\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\left(1\right)\\\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\left(2\right)\)
Từ (1),(2) suy ra \(\frac{x}{9}=\frac{y}{12}=\frac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{9}=\frac{y}{12}=\frac{z}{15}=\frac{2x}{18}=\frac{-3y}{-36}=\frac{z}{15}=\frac{2x-3y+z}{18-\left(-36\right)+15}=\frac{6}{69}=\frac{2}{23}\)Suy ra x =\(\frac{2}{23}\cdot9=\frac{18}{23}\)
\(y=\frac{2}{23}\cdot12=\frac{24}{23}\\ z=\frac{2}{23}.15=\frac{30}{23}\)
a) Điều kiện: x > 0
\(\sqrt{\left(1-2x\right)^2}=x\) => (1 - 2x)2 = x2 => 1 - 2x = x hoặc 1 - 2x = - x
+) 1 - 2x = x => 1 = 3x => x = 1/3 (Thỏa mãn)
+) 1 - 2x = - x => 1 = x (Thỏa mãn)
Vậy....
b) \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)
=> \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\)
=> \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)\left(x+1\right)=0\)
=> \(\frac{43}{60}\left(x+1\right)=0\)=> x + 1 = 0 => x = - 1
Vậy....
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
a)\(\left(2x-3\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\begin{cases}2x-3>0\\x+1< 0\end{cases}\) hoặc \(\begin{cases}2x-3< 0\\x+1>0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{3}{2}\\x< -1\end{cases}\) (loại) hoặc \(\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)
\(\Leftrightarrow-1< x< \frac{3}{2}\)
b) \(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)
\(\Leftrightarrow\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{2}\\x< -3\end{array}\right.\)
c) Sai đề phải là \(\frac{x}{\left(x+3\right)\left(x+7\right)}\)
Có: \(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+3\right)\left(x+17\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow x=4\)
\(a,\frac{x+2}{x-3}=\frac{x-6}{x+11}\)
\(\Rightarrow\left(x+2\right)\left(x+11\right)=\left(x-3\right)\left(x-6\right)\)
\(\Leftrightarrow x^2+13x+22=x^2-9x+18\)
\(\Leftrightarrow22x=-4\)
\(\Rightarrow x=-\frac{2}{11}\)
a) Đề
=> ( x + 2 ) ( x + 11 ) = ( x - 3) ( x - 6 )
<=> x2 + 13x + 22 = x2 - 9x + 18
<=> 22x = -4
<=> x = - 2 /11
b) Đề
=> ( 2x + 5 ) ( 2x -3 ) = ( x -1) ( 4x - 6 )
<=> 4x2 + 4x - 15 = 4x2 - 10x + 6
<=> 14x = 21
<=> x = 21/14