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a) \(2x+\frac{3}{15}=\frac{7}{5}\)
=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)
=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)
b) \(x-\frac{2}{9}=\frac{8}{3}\)
=> \(x=\frac{8}{3}+\frac{2}{9}\)
=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)
c) \(\frac{-8}{x}=\frac{-x}{18}\)
=> x(-x) = (-8).18
=> -x2 = -144
=> x2 = 144(bỏ dấu âm)
=> x = \(\pm\)12
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)
=> 5(2x + 3) = 6(x - 2)
=> 10x + 15 = 6x - 12
=> 10x + 15 - 6x + 12 = 0
=> 4x + 27 = 0
=> 4x = -27
=> x = -27/4
e) \(\frac{x+1}{22}=\frac{6}{x}\)
=> x(x + 1) = 132
=> x(x + 1) = 11.12
=> x = 11
f) \(\frac{2x-1}{2}=\frac{5}{x}\)
=> x(2x - 1) = 10
=> 2x2 - x = 10
=> 2x2 - x - 10 = 0
tới đây tự làm đi nhé
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
=> (2x - 1)(2x + 1) = 63
=> 4x2 - 1 = 63
=> 4x2 = 64
=> x2 = 16
=> x = \(\pm\)4
h) Tương tự
a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)
b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)
c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)
e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
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\(a,\frac{8}{x}=\frac{x}{4}\)
\(=>x\cdot x=8\cdot4\)
\(=>x^2=32\)
\(=>x=\sqrt{32}\)
\(c,\frac{2x+3}{6}=\frac{x+1}{-8}\)
\(=>-8\cdot\left(2x+3\right)=6\cdot\left(x+1\right)\)
\(=>-16x-24=6x+6\)
\(=>-16x-6x=6+24\)
\(=>-22x=30\)
\(=>x=\frac{30}{-22}=-\frac{15}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)
\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)
\(\Rightarrow27x+15=96\)
\(\Rightarrow27x=81\)
\(\Rightarrow x=3\left(tm\right)\)
\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\left(tm\right)\)
#Toru
a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)
\(\Rightarrow-6x+8x+3x+3+4x+2=32\)
\(\Rightarrow9x+5=32\)
\(\Rightarrow9x=32-5\)
\(\Rightarrow9x=27\)
\(\Rightarrow x=\dfrac{27}{9}\)
\(\Rightarrow x=3\)
b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\))
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=\dfrac{12}{2}\)
\(\Rightarrow x=6\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(\frac{6^{x+3}-6^{x+1}+6^x}{211}=\frac{7^{2x}+7^{2x+1}+7^{2x-3}}{8\frac{1}{49}}\)
\(\Leftrightarrow \frac{6^x(6^3-6+1)}{211}=\frac{7^{2x}(1+7+\frac{1}{7^3})}{\frac{393}{49}}\)
\(\Leftrightarrow 6^x=7^{2x}.\frac{915}{917}\)
\(\Leftrightarrow (\frac{6}{49})^x=\frac{915}{917}\)
\(\Rightarrow x=\log_{\frac{6}{49}}\frac{915}{917}\)
Trần Linh: cách giải này gây khó hiểu cho bạn ở dòng cuối đúng không? Nếu không dùng log thì không thể tìm ra kết quả cuối cùng theo cách lớp 7 do nghiệm quá xấu. Do đó, bạn hãy xem lại đề xem có nhầm dấu hay viết sai ở chỗ nào không.
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a) \(\left|4-x\right|+2x=3\)
<=> \(\left|4-x\right|=3-2x\)
<=> \(\orbr{\begin{cases}4-x=3-2x\left(x\le4\right)\\x-4=3-2x\left(x>4\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\left(tm\right)\\3x=7\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\\x=\frac{7}{3}\left(ktm\right)\end{cases}}\)
Vậy x = -1
b) \(\left|x-7\right|+2x+5=6\)
<=> \(\left|x-7\right|=1-2x\)
<=> \(\orbr{\begin{cases}x-7=1-2x\left(đk:x\ge7\right)\\x-7=2x-1\left(đk:x< 7\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=8\\x=-6\left(tm\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{8}{3}\left(ktm\right)\\x=-6\left(tm\right)\end{cases}}\)
Vậy x = -6
c) \(3x-\left|2x+1\right|=2\)
<=> \(\left|2x+1\right|=3x-2\)
<=> \(\orbr{\begin{cases}2x+1=3x-2\left(đk:x\ge-\frac{1}{2}\right)\\2x+1=2-3x\left(đk:x< -\frac{1}{2}\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\left(tm\right)\\5x=1\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{5}\left(ktm\right)\end{cases}}\)
Vậy x = 3
d) \(\left|x+2\right|-x=2\)
<=> \(\left|x+2\right|=x+2\)
<=> \(\orbr{\begin{cases}x+2=x+2\left(đk:x\ge-2\right)\\x+2=-x-2\left(x< -2\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}0x=0\\2x=-4\end{cases}}\)
<=> 0x = 0 (luôn đúng) và x = -2 (ktm)
Vậy x \(\ge\)-2
e) \(\left|x-3\right|=21\)
<=> \(\orbr{\begin{cases}x-3=21\\3-x=21\end{cases}}\)
<=> \(\orbr{\begin{cases}x=24\\x=-18\end{cases}}\)
Vậy x = 24 hoặc x = -18
f) \(\left|2x+3\right|-\left|x-3\right|=0\)
<=> \(\left|2x+3\right|=\left|x-3\right|\)
<=> \(\orbr{\begin{cases}2x+3=x-3\\2x+3=3-x\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\3x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=0\end{cases}}\)
Vậy x thuộc {-6; 0}
g) Ta có: \(\left|x+\frac{1}{8}\right|\ge0\forall x\)
\(\left|x+\frac{2}{8}\right|\ge0\forall x\)
\(\left|x+\frac{5}{8}\right|\ge0\forall x\)
=> VT = \(\left|x+\frac{1}{8}\right|+\left|x+\frac{2}{8}\right|+\left|x+\frac{5}{8}\right|\ge0\forall x\)
=> VP \(\ge0\) => \(4x\ge0\) => \(x\ge0\)
Do đó: \(x+\frac{1}{8}+x+\frac{2}{8}+x+\frac{5}{8}=4x\)
<=> \(3x+1=4x\) <=> \(x=1\left(tm\right)\)
Vậy x = 1
h) \(\left|x-2\right|-\left|2x+3\right|-x=-2\)
<=> \(\left|x-2\right|-\left|2x+3\right|=x-2\)(*)
Lập bảng xét dấu:
x -3/2 2
x - 2 2 - x | 2 - x 0 x - 2
2x + 3 -2x - 3 0 2x + 3 | 2x + 3
Xét x < -3/2 => pt (*) trở thành: 2 - x + 2x + 3 = x - 2
<=> x + 5 = x - 2 <=> 0x = -7 (vô lí)
Xét -3/2 \(\le\) x < 2 => pt (*) trở thành: 2 - x - 2x - 3 = x - 2
<=> 4x = 1 <=> x = 1/4 ((tm)
Xét x \(\ge\) 2 => pt (*) trở thành x - 2 - 2x - 3 = x - 2
<=> 2x = -3 <=> x = -3/2 (ktm)
Vậy x = 1/4
i) |2x - 3| - x = |2 - x|
<=> |2x - 3| - |2 - x| = x (*)
Lập bảng xét dấu
x 3/2 2
2x - 3 3 - 2x 0 2x - 3 | 2x - 3
2 - x 2 - x | 2 - x 0 x - 2
Xét x < 3/2 => pt (*) trở thành: 3 - 2x - 2 + x = x
<=> 2x = 1 <=> x = 1//2 ((tm)
Xét \(\frac{3}{2}\le x< 2\)=> pt (*) trở thành: 2x - 3 - 2 + x = x
<=> 2x = 5 <=> x = 5/2 (ktm)
Xét x \(\ge\)2 ==> pt (*) trở thành: 2x - 3 - x + 2 = x
<=> 0x = -5 (vô lí)
Vậy x = 1/2
k) 2|x - 3| - |4x - 1| = 0
<=> 2|x - 3| = |4x - 1|
<=> \(\orbr{\begin{cases}2\left(x-3\right)=4x-1\\2\left(x-3\right)=1-4x\end{cases}}\)
<=> \(\orbr{\begin{cases}2x-6=4x-1\\2x-6=1-4x\end{cases}}\)
<=> \(\orbr{\begin{cases}2x=-5\\6x=7\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{5}{2}\\x=\frac{7}{6}\end{cases}}\) Vậy ...
Nếu x < -3
=> |x + 3| = -(x + 3) = -x - 3
=> |2x - 6| = -(2x - 6) = -2x + 6
Khi đó |2x -6| + |x + 3| = 8 (1)
<=> -2x + 6 - x - 3 = 8
=> -3x + 3 = 8
=> -3x = 5
=> x = -5/3 (loại)
Nếu -3 \(\le x\le3\)
=> |2x - 6| = -(2x - 6) = -2x + 6
=> |x + 3| = x + 3
Khi đó (1) <=> - 2x + 6 + x + 3= 8
=> -x = -1
=> x = 1(tm)
Khi x > 3
=> |2x - 6| = 2x - 6
=> |x + 3| = x + 3
Khi đó (1) <=> 2x - 6 + x + 3 = 8
=> 3x - 3 = 8
=> 3x = 11
=> x = 11/3 (tm)
Vậy \(x\in\left\{1;\frac{11}{3}\right\}\)là giá trị cần tìm