|x-10|10+|x-11|11=11
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\(\dfrac{9}{10}\times\left(\dfrac{11}{29}+\dfrac{2}{29}+\dfrac{16}{29}\right)=\dfrac{9}{10}\times\dfrac{29}{29}=\dfrac{9}{10}\times1=\dfrac{9}{10}\)
\(\frac{x+5}{13}+\frac{x+6}{12}+\frac{x+7}{11}=\frac{x+8}{10}+\frac{x+9}{9}+\frac{x+10}{8}\)
\(\Leftrightarrow\left(\frac{x+5}{13}+1\right)+\left(\frac{x+6}{12}+1\right)+\left(\frac{x+7}{11}+1\right)=\left(\frac{x+8}{10}+1\right)+\left(\frac{x+9}{9}+1\right)+\left(\frac{x+10}{8}\right)\)
\(\Leftrightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}=\frac{x+18}{10}+\frac{x+18}{9}+\frac{x+18}{8}\)
ta chuyển về vế trái được
\(\Leftrightarrow\left(x+18\right)\left(\frac{1}{13}+\frac{1}{122}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)
\(\Leftrightarrow x+2018=0\)(do cái còn lại khác 0)
\(\Leftrightarrow x=-2018\)
mình nghĩ đề cậu viết thiếu mình sửa rồi
Ta có:
\(\frac{x+5}{13}+\frac{x+6}{12}+\frac{x+7}{11}=\frac{x+8}{10}+\frac{x+9}{9}+\frac{x+10}{8}\)
\(\Rightarrow\left(\frac{x+5}{13}+1\right)+\left(\frac{x+6}{12}+1\right)+\left(\frac{x+7}{11}+1\right)=\left(\frac{x+8}{10}+1\right)+\left(\frac{x+9}{9}+1\right)+\left(\frac{x+10}{8}+1\right)\)
\(\Rightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}=\frac{x+18}{10}+\frac{x+18}{9}+\frac{x+18}{8}\)
\(\Rightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}-\frac{x+18}{10}-\frac{x+18}{9}-\frac{x+18}{8}=0\)
\(\Rightarrow\left(x+18\right)\times\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)
Vì \(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\ne0\)
\(\Rightarrow x+18=0\)
\(\Rightarrow x=-18\)
Vậy phương trình có nghiệm là x = -18
Bài làm
m) (x + 2).(3 - x) = 0;
=> x + 2 = 0 hoặc 3 - x = 0
=> x = -2 hoặc x = 3
Vậy x = -2 hoặc x = 3
d) 511.712 + 511.711
= 511 . ( 712 + 711 )
= 511 . [ 711 . ( 7 + 1 ) ]
= 511 . 711 . 8
= ( 5 . 7 )11 . 8
= 3511 . 8
512.712 + 9.511.711
= 511 ( 5 . 712 + 9 . 1 . 711 )
= 511 [ 711 ( 5 . 7 + 9 . 1 . 1 ) ]
= 511 ( 711 . 44 )
= 511 . 711 . 44
= 3511 . 44
m. \(\left(x+2\right)\left(3-x\right)=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\3-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
d. \(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.\left(7^{12}+7^{11}\right)}{5^{11}.\left(5.7^{12}+9.7^{11}\right)}=\frac{7^{12}+7^{11}}{5.7^{12}+9.7^{11}}=\frac{1}{5.9}=\frac{1}{45}\)
q. \(\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10+11=11\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10=0\)
\(\Rightarrow\left[\left(x-3\right)+\left(x-2\right)+\left(x-1\right)\right]+(1+2+3+...+10)=0\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+55=0\)
\(\Rightarrow x-3+x-2+x-1=-55\)
\(\Rightarrow3x-6=-55\)
\(\Rightarrow3x=-49\)
\(\Rightarrow x=-\frac{49}{3}\)
[6: 3/5 - 7/6 x 6/7] : [ 21/5 x 10/11 + 57/11]
= 9 : 9
= 1
\(\left(6:\frac{3}{5}-\frac{7}{6}\times\frac{6}{7}\right):\left(\frac{21}{5}\times\frac{10}{11}+\frac{57}{11}\right)\)
\(=\left(10-1\right):\left(\frac{42}{11}+\frac{57}{11}\right)\)
\(=9:9\)
\(=1\)
a. $\dfrac{14}{15}+\dfrac{2}{11}+\dfrac{1}{15}+\dfrac{4}{11}+\dfrac{10}{22}$
$=(\dfrac{14}{15}+dfrac{1}{15})+(\dfrac{2}{11}+\dfrac{4}{11}+\dfrac{10}{22})$
$=1+1=2$
b. $74\times43-43+27\times43$
$=(74-1+27)\times43$
$=100\times43=4300$
\(a)\dfrac{14}{15}+\dfrac{2}{11}+\dfrac{1}{15}+\dfrac{4}{11}+\dfrac{10}{22}\)
\(=\left(\dfrac{14}{15}+\dfrac{1}{15}\right)+\dfrac{2}{11}+\dfrac{4}{11}+\dfrac{5}{11}\)
\(=1+1=2\)
\(b)74\times43-43+27\times43\)
\(=43\times\left(74-1+27\right)\)
\(=43\times100=4300\)
6/20:3=6/20x1/3=6/60=1/10
11h-8h30p=2h30p=2.5h
10 7/10-4 3/10
=107/10-43/10
=64/10=32/5
chúc bn học tốt!
11) \(\sqrt{\dfrac{3}{4}x}-\dfrac{1}{2}=\dfrac{3}{7}\left(x\ge0\right)\)
\(\Rightarrow\sqrt{\dfrac{3}{4}x}=\dfrac{3}{7}+\dfrac{1}{2}\)
\(\Rightarrow\sqrt{\dfrac{3}{4}x}=\dfrac{13}{14}\)
\(\Rightarrow\dfrac{3}{4}x=\left(\dfrac{13}{14}\right)^2\)
\(\Rightarrow\dfrac{3}{4}x=\dfrac{169}{196}\)
\(\Rightarrow x=\dfrac{169}{196}:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{169}{147}\)
12) \(\dfrac{2}{3}+\sqrt{\dfrac{1}{3}:x}=\dfrac{3}{5}\left(x>0\right)\)
\(\Rightarrow\sqrt{\dfrac{1}{3}:x}=\dfrac{3}{5}-\dfrac{2}{3}\)
\(\Rightarrow\sqrt{\dfrac{1}{3}:x}=-\dfrac{1}{15}\)
Do biểu thức trong dấu căn luôn dương nên không có x thỏa mãn