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a) Đặt \(x-1=a\)
\(pt\Leftrightarrow\frac{13}{a}+\frac{5}{2a}=\frac{6}{3a}\)
\(\Leftrightarrow\frac{31}{2a}=\frac{6}{3a}\)
\(\Leftrightarrow\frac{31}{2}=2\)(vô lí)
Vậy pt vô nghiệm
a) \(\frac{13}{x-1}+\frac{5}{2x-2}=\frac{6}{3x-3}\)
\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{6}{3\left(x-1\right)}\)
\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{2}{x-1}\)
\(\frac{31}{2\left(x-1\right)}=\frac{2}{x-1}\)
\(\frac{31}{2}=2\)
=> không có x thỏa mãn đề bài.
b) \(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
\(\frac{1}{x-1}+\frac{-2}{3}.\frac{-9}{20}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}-\frac{-18}{60}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2\left(1-x\right)}\)
\(10\left(1-x\right)+3\left(x-1\right)\left(1-x\right)=25\left(x-1\right)\)
\(7-4x-3x^2=25x-25\)
\(7-4x-3x^2-25x+25=0\)
\(32-29x-3x^2=0\)
\(3x^2+29x-30=0\)
\(3x^2+32x-3x-32=0\)
\(x\left(3x+32\right)-\left(3x+32\right)=0\)
\(\left(3x+32\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}3x+32=0\\x-1=0\end{cases}}\)
\(\orbr{\begin{cases}x=-\frac{32}{3}\\x=1\end{cases}}\)
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 ~
\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=8^{\left|2x+6\right|}\)
\(\Leftrightarrow\frac{4.\left(4^5\right)}{3.\left(3^5\right)}.\frac{6.\left(6^5\right)}{2.\left(2^5\right)}=8^{\left|2x+6\right|}\)
\(\Leftrightarrow\frac{4^6.6^6}{3^6.2^3}=8^{\left|2x+6\right|}\)
\(\Leftrightarrow\frac{\left(2^2\right)^6.\left(2.3\right)^6}{3^6.2^3}=8^{\left|2x+6\right|}\)
\(\frac{2^{12}.2^6.3^6}{3^6.2^3}=\frac{2^{18}.3^6}{3^6.2^3}=\frac{2^{15}.1}{1.1}=2^{15}=8^{\left|2x+6\right|}\)
=> 215=(23)|2x+6|
215=23|2x+6|
<=> 3|2x+6|=15
|2x+6|=15:3
|2x+6|=5
\(\Rightarrow\orbr{\begin{cases}2x+6=5\\2x+6=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{-11}{2}\end{cases}}\)
1) \(\frac{x-1}{x-5}=\frac{6}{7};\left(x-1\right).7=\left(x-5\right).6\)
7x - 7 = 6x - 30
=> 7x - 6x = -30 - (-7)
x = -23
2) \(\frac{x-1}{3}=\frac{x+3}{5};\left(x-1\right).5=\left(x+3\right).3\)
5x - 5 = 3x + 9
=> 5x - 3x = 9 - (-5)
2x = 14
x = 7
3) \(\frac{3}{7}=\frac{2x+1}{3x+5};\left(3x+5\right).3=\left(2x+1\right).7\)
9x + 15 = 14x + 7
9x - 14x = 7-15
5x = -8
x = -8/5
1) =>\(\hept{\begin{cases}x-1=6\\x-5=7\end{cases}=>\hept{\begin{cases}x=6+1=7\\x=7+5=13\end{cases}}}\)
Vậy x\(\varepsilon\){7;13}
2)
\(2\left|2x-6\right|=\frac{5}{6}-\left|x-3\right|\)
\(\Rightarrow\left|4x-12\right|=\frac{5}{6}-x+3\)
\(\Rightarrow4x-12=\frac{5}{6}+\frac{18}{6}-x\)
\(\Rightarrow4x-12=\frac{23}{6}-x\)
Từ đây chuyển vế đổi dấu là xong nhé
Lập bảng xét dấu...! :(