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Lời giải: Giải phương trình với tập xác định
Tập xác định của phương trình
\(x\in\infty-\infty\)
\(\frac{19x+67}{90}=\frac{15x+83}{56}\Rightarrow\left(19x=67\right)56=90\left(15x+83\right)\)
Kết quả : \(-13\)
kq đúng nhưng mk k biết mấy cái phương trình đó vì mk mới lớp 7
\(\frac{x+1}{10}+\frac{x+2}{9}+\frac{x+3}{8}+\frac{x+4}{7}+\frac{x+5}{6}=-5\)
\(\left(\frac{x+1}{10}+1\right)+\left(\frac{x+2}{9}+1\right)+\left(\frac{x+3}{8}+1\right)+\left(\frac{x+4}{7}+1\right)+\left(\frac{x+5}{6}+1\right)=0\)
\(\frac{x+11}{10}+\frac{x+11}{9}+\frac{x+11}{8}+\frac{x+11}{7}+\frac{x+12}{6}=0\)
\(\left(x+11\right)\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+\frac{1}{6}\right)=0\)
Vì : \(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+\frac{1}{6}>0\)
\(\Rightarrow x+11=0\)
\(\Rightarrow x=-11\)
Câu b) tạm thời ko bít làm =.=
Bài 1 :
\(d)\) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)
\(\Leftrightarrow\)\(\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=2x\)
\(\Leftrightarrow\)\(\frac{4^6}{3^6}.\frac{6^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{2^6.3^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{3^6}{1}=2x\)
\(\Leftrightarrow\)\(2^{12}=2x\)
\(\Leftrightarrow\)\(x=\frac{2^{12}}{2}\)
\(\Leftrightarrow\)\(x=2^{11}\)
\(\Leftrightarrow\)\(x=2048\)
Vậy \(x=2048\)
Chúc bạn học tốt ~
Bài 1 :
\(a)\) Ta có :
\(4+\frac{x}{7+y}=\frac{4}{7}\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{4}{7}-4\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{-24}{7}\)
\(\Leftrightarrow\)\(\frac{x}{-24}=\frac{7+y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{-24}=\frac{7+y}{7}=\frac{x+7+y}{-24+7}=\frac{22+7}{-17}=\frac{29}{-17}=\frac{-29}{17}\)
Do đó :
\(\frac{x}{-24}=\frac{-29}{17}\)\(\Rightarrow\)\(x=\frac{-29}{17}.\left(-24\right)=\frac{696}{17}\)
\(\frac{7+y}{7}=\frac{-29}{17}\)\(\Rightarrow\)\(y=\frac{-29}{17}.7-7=\frac{-322}{17}\)
Vậy \(x=\frac{696}{17}\) và \(y=\frac{-322}{17}\)
Chúc bạn học tốt ~
a, \(\left|x+\frac{1}{3}\right|=0\Leftrightarrow x=-\frac{1}{3}\)
b, \(\left|\frac{5}{18}-x\right|-\frac{7}{24}=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{18}-x=\frac{7}{24}\\\frac{5}{18}-x=-\frac{7}{24}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{72}\\x=\frac{41}{72}\end{cases}}\)
c, \(\frac{2}{5}-\left|\frac{1}{2}-x\right|=6\Leftrightarrow\left|\frac{1}{2}-x\right|=-\frac{28}{5}\)vô lí
Vì \(\left|\frac{1}{2}-x\right|\ge0\forall x\)*luôn dương* Mà \(-\frac{28}{5}< 0\)
=> Ko có x thỏa mãn
\(|x+\frac{1}{3}|=0\)
\(< =>x+\frac{1}{3}=0< =>x=-\frac{1}{3}\)
\(|x+\frac{3}{4}|=\frac{1}{2}\)
\(< =>\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)
a, ( 152 +và 2/4 - 148 và 3/8 ) : 0,2 = x : 0,3
=> 33/8 : 1/5 = x : 3/10
=> x : 3/10 = 165/8
=> x = 99/10
b, ( 85 và 7/30 - 83 và 5/18 ) : 2 và 2/3 = 0,01x : 4
=> 88/45 : 8/3 = 0,01x : 4
=> 0,01x : 4 = 11/15
=> 0,01x = 44/15
=> x = 880/3
c, x - 1/ x + 5 = 6/7
=> 7( x - 1 ) = 6( x + 5 )
=> 7x - 7 = 6x + 30
=> 7x - 6x = 7 + 30
=> x = 37
d, x2/6 = 24/25
=> x2. 25 = 6 . 24
=> x2.25 = 144
=> x2 = 144/25
=> x = ( 12/5)2 hoặc x = ( -12/5)
g, x - 3/ x + 5 = 5/7
=> 7( x - 3 ) = 5 ( x + 5 )
=> 7x - 21 = 5x + 25
=> 7x - 5x = 21 + 25
=> 2x = 46
=> x = 23
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)
Vậy \(x = \frac{{ - 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ - 9}}{{10}}\).
\(\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}\)
\(\Rightarrow\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}+3=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}+3\)
\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+4}{6}+1\right)+\left(\frac{x+5}{5}+1\right)=\left(\frac{x+2}{8}+1\right)\)\(+\left(\frac{x+3}{7}+1\right)+\left(\frac{x+6}{4}\right)\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}\right)=\left(x+10\right)\left(\frac{1}{8}+\frac{1}{7}+\frac{1}{4}\right)\)
\(\Rightarrow\left(x+10\right)\frac{43}{90}=\left(x+10\right)\frac{29}{56}\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\)
cộng 3 vào cả hai vế nên phương trình vẫn bằng nhau
Ta có \(\frac{x+1}{9}+1+\frac{x+4}{6}+1+\frac{x+5}{5}+1=\frac{x+2}{8}+1+\frac{x+3}{7}+1+\frac{x+6}{4}+1\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}-\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{4}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
mà \(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)
\(\Rightarrow x+10=0\)
\(\Leftrightarrow x=-10\)