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Có:
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Leftrightarrow\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}-\dfrac{x+2}{14}-\dfrac{x+2}{15}=0\)
\(\Leftrightarrow\left(x+2\right)\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)
Dấu "=" xảy ra:
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}=0\end{matrix}\right.\)
Vì \(\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)\ne0\)
\(\Leftrightarrow x-2=0\)
\(\Rightarrow x=0+2=2\)
Vậy \(x=2\).
Học tốt!
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\left(\dfrac{1}{11}+\dfrac{1}{12}\right)\left(x+2\right)+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\dfrac{23\left(x+2\right)}{132}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\left(\dfrac{23}{132}+\dfrac{1}{13}\right)\left(x+2\right)=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\left(\dfrac{1}{14}+\dfrac{1}{15}\right)\left(x+2\right)\)
\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\dfrac{29\left(x+2\right)}{210}\)
\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}-\dfrac{29\left(x+2\right)}{210}=0\)
\(\Rightarrow\left(\dfrac{431}{6.286}-\dfrac{29}{6.35}\right)\left(x+2\right)=0\)
\(\Rightarrow\dfrac{1}{6}\left(\dfrac{431}{286}-\dfrac{29}{35}\right)\left(x+2\right)=-2\)
x-2/11 + x-2/12 +x-2/13 = x-2/14 + x-2/15
=> x-2 /11 + x-2/12 +x-2/13 - x-2/14 - x-2/15 = 0
=> (x-2). ( 1/11 + 1/12 + 1/13 - 1/14-1/15) = 0
=> x-2 = 0 => x=2
1/ 11 + 1/12 +1/13 -1/14 - 1/15 = 0
Vì 1/11; 1/12; 1/13; 1/14; 1/15 > 1 nên 1/11+1/12+1/3-1/14-1/15= 0 (vô lí)
Vậy x=2
Nhớ like
a, \(\left(x-1\right)^5=-243\)
=> \(\left(x-1\right)^5=\left(-3\right)^5\)
=> x-1= -3
=> x= -2
b, \(\dfrac{x+2}{11}+\dfrac{2+x}{12}+\dfrac{x+2}{13}=\dfrac{2+x}{14}+\dfrac{x+2}{15}\)
=> \(\dfrac{x+2}{11}+\dfrac{2+x}{12}+\dfrac{x+2}{13}-\dfrac{2+x}{14}+\dfrac{x+2}{15}=0\)
=>\(\dfrac{x+2+2+x+x+2-2+x+x+2}{11+12+13-14+15}\)
=> \(\dfrac{x+2}{37}=0\)
=> x+2= 0
=> x=-2
a, \(\left(x-1\right)^5=-243\)
\(\Leftrightarrow\left(x-1\right)^5=-3^5\)
\(\Leftrightarrow x-1=-3\Leftrightarrow x=-2\)
b,\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}-\dfrac{x+2}{14}-\dfrac{x+2}{15}=0\)
\(\Leftrightarrow\left(x+2\right).\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)
\(do\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\ne0\)
\(\Rightarrow x+2=0\Leftrightarrow x=-2\)
c, \(x-2\sqrt{x}=0\Leftrightarrow\sqrt{x^2}-2\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\sqrt{2}\end{matrix}\right.\)
a) Ta có:
(x - 1)5 = - 243
=> (x - 1)5 = (-3)5
=> x - 1 = - 3
=> x = -3 + 1
=> x = -2
Vậy x = -2
b) Ta có:
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\left(x+2\right).\dfrac{1}{11}+\left(x+2\right).\dfrac{1}{12}+\left(x+2\right).\dfrac{1}{13}=\left(x+2\right).\dfrac{1}{14}+\left(x+2\right).\dfrac{1}{15}\)
=> \(\left(x+2\right).\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}\right)=\left(x+2\right).\left(\dfrac{1}{14}+\dfrac{1}{15}\right)\)
=> \(\left(x+2\right).\dfrac{431}{1716}=\left(x+2\right).\dfrac{29}{210}\)
=> \(\left(x+2\right).\dfrac{431}{1716}-\left(x+2\right).\dfrac{29}{210}=0\)
=> (x + 2).(\(\dfrac{431}{1716}-\dfrac{29}{210}\)) = 0
mà \(\dfrac{431}{1716}-\dfrac{29}{210}\) \(\ne\) 0
=> x + 2 = 0
=> x = -2
Vậy x = -2
c) Ta có :
\(\left|3x-2\right|+5x=4x-10\)
=> \(\left|3x-2\right|=4x-5x-10\)
=> \(\left|3x-2\right|=-x-10\)
=> 3x - 2 = -x - 10
hoặc 3x - 2 = -(-x -10)
*) Nếu 3x - 2 = -x - 10
=> 3x + x = -10 + 2
=> 4x = -8
=> x = -2
*) Nếu 3x - 2 = -(-x -10)
=> 3x - 2 = x +10
=> 3x - x = 10 + 2
=> 2x = 12
=> x = 6
Vậy x = -2 hoặc x = 6
a) Ta có : \(x - 2xy + y - 3 = 0\)
\(\Rightarrow-2xy+x+y=3\)
\(\Rightarrow-2.\left(-2xy+x+y\right)=-2.3\)
\(\Rightarrow4xy-2x-2y=-6\)
\(\Rightarrow4xy-2x-2y+1=-6+1\)
\(\Rightarrow2x.\left(2y-1\right).\left(2y-1\right)=-5\)
\(\Rightarrow\left(2y-1\right).\left(2x-1\right)=-5=1.\left(-5\right)=-5.1=\left(-1\right).5=5.\left(-1\right)\)
Tự lập bảng đi -.-
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Nhân từng vế bất đẳng thức ta được : (xyz)2 = 36xyz + Nếu một trong các số x,y,z bằng 0 thì 2 số còn lại cũng bằng 0 + Nếu cả 3 số x,y,z khác 0 thì chia 2 vế cho xyz ta được xyz = 36 + Từ xyz =36 và xy = z ta được z2 = 36 nên z = 6; z = -6 + Từ xyz =36 và yz = 4x ta được 4x2 = 36 nên x = 3; x = -3 + Từ xyz =36 và ta được 9y2 = 36 nên y = 2; y = -2 - Nếu z = 6 thì x và y cùng dấu nên x = 3, y = 2 hoặc x = -3 , y = -2 - Nếu z = -6 thì x và y trái dấu nên x = 3 ; y = -2 hoặc x = -3; y=2 |
Vậy có 5 bộ số (x, y, z) thoã mãn: (0,0,0); (3,2,6);(-3,-2,6);(3,-2,-6);(-3,2.-6)
\(1,\)
\(a,\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)\)
\(=\dfrac{11}{125}+\left(\dfrac{-1}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{11}{125}\)
\(b,-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=\dfrac{-12}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=-15.\left[\dfrac{12}{7}+\dfrac{2}{7}+\left(-5\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\right]\)
\(=-15.\left[2+\left(-5\right).\dfrac{1}{105}\right]\)
\(=-15.\left(2-\dfrac{1}{21}\right)\)
\(=-15.\dfrac{41}{21}=\dfrac{-615}{21}\)
\(2,\)
\(a,\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
\(\Leftrightarrow\dfrac{11}{13}-\dfrac{5}{42}+x=\dfrac{-15}{28}+\dfrac{11}{13}\)
\(\Leftrightarrow x=\dfrac{-15}{28}+\dfrac{11}{13}-\dfrac{11}{13}+\dfrac{5}{42}\)
\(\Leftrightarrow x=\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\left(\dfrac{5}{42}+\dfrac{-15}{28}\right)\)
\(\Leftrightarrow x=\dfrac{5}{12}\)
Vậy \(x=\dfrac{5}{12}\)
\(b,\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6=\dfrac{16}{10}=\dfrac{8}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=\dfrac{-8}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{5}-\dfrac{4}{15}=\dfrac{4}{3}\\x=\dfrac{-8}{5}-\dfrac{4}{15}=\dfrac{-28}{15}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{4}{3};\dfrac{-28}{15}\right\}\)
\(c,7^{x+2}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.\left(7^3+2\right)=345\)
\(\Leftrightarrow7^{x-1}.\left(343+2\right)=345\)
\(\Leftrightarrow7^{x-1}.345=345\)
\(\Leftrightarrow7^{x-1}=345:345=1\)
\(\Leftrightarrow x-1=0\)
\(x=0+1=1\)
Vậy \(x=1\)
Các câu dễ tự làm :v
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
Giải:
Ta có:
\(\dfrac{x-2}{11}+\dfrac{x-2}{12}+\dfrac{x-2}{13}=\dfrac{x-2}{14}+\dfrac{x-2}{15}\)
\(\Leftrightarrow\dfrac{x-2}{11}+\dfrac{x-2}{12}+\dfrac{x-2}{13}-\dfrac{x-2}{14}-\dfrac{x-2}{15}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)
Vì \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\ne0\)
Nên \(x-2=0\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\).
Chúc bạn học tốt!