TÌM X:\(\frac{3X+2}{X-1}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt bt trên là A nha
Đổi |x-1|=|1-x|
Suy ra A=|1-x|+x-2|+|x-3|
Áp dụng BĐTGTTĐ ta có
A=|1-x|+x-2|+|x-3|\(\ge\)|1-x+x-3|=2
Dấu = xảy ra khi \(\hept{\begin{cases}x-2=0\\1< x< 3\end{cases}}\)đồng thời xảy ra
Vậy x =2
b,
\(\left|3x+\frac{1}{2}\right|\ge0\)
\(\left|3x+\frac{1}{6}\right|\ge0\)
..........
\(\left|3x+380\right|\ge0\)
Suy ra đề bài \(\ge\)0
suy ra 58x \(\ge\)0
Suy ra \(3x+\frac{1}{2}+3x+\frac{1}{6}+......+3x+380=58x\)
Tự tính nhé hok tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(ĐKXĐ:\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(=\left[\frac{\left(x+1\right)\left(x+2\right)}{3x\left(x+1\right)}+\frac{6x}{3x\left(x+1\right)}-\frac{9x\left(x+1\right)}{3x\left(x+1\right)}\right]:\frac{2\left(1-2x\right)}{x+1}-\frac{3x+1-x^2}{3x}\)
\(=\frac{\left(x+1\right)\left(x+2\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(=\frac{2-8x^2}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(=\frac{1+2x-3x-1+x^2}{3x}\)
\(=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)
b)\(\text{Với }x\ne0,x\ne-1,x\ne\frac{1}{2}\text{ ta có:}\)
\(\text{Để A< 0\Leftrightarrow}\frac{x-1}{3}< 0\Rightarrow x-1< 0\Leftrightarrow x< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
pt \(\Leftrightarrow\left(2^x\right)^3-\left(\frac{1}{2^{x-1}}\right)^3-6\left(2^x-\frac{1}{2^{x-1}}\right)=1\)
\(\Leftrightarrow\left(2^x-\frac{1}{2^{x-1}}\right)\left(2^{2x}+2^x\cdot\frac{1}{2^{x-1}}+\left(\frac{1}{2^{x-1}}\right)^2\right)-6\left(2^x-\frac{1}{2^{x-1}}\right)=1\)
\(\Leftrightarrow\left(2^x-\frac{1}{2^{x-1}}\right)\left(2^{2x}+\frac{1}{2^{2x-2}}-4\right)=1\)
\(\Leftrightarrow\frac{2^{2x-1}-1}{2^{x-1}}\cdot\frac{2^{4x-2}-4\cdot2^{2x-2}+1}{2^{2x-1}}=1\)
\(\Leftrightarrow\frac{2^{2x-1}-1}{2^{x-1}}\cdot\frac{2^{2\left(2x-1\right)}-2\cdot2^{2x-1}+1}{2^{2x-2}}=1\)
\(\Leftrightarrow\frac{2^{2x-1}-1}{2^{x-1}}\cdot\left(\frac{2^{2x-1}-1}{2^{x-1}}\right)^2=1\)
\(\Leftrightarrow\left(\frac{2^{2x-1}-1}{2^{x-1}}\right)^3=1\)
\(\Leftrightarrow\frac{2^{2x-1}-1}{2^{x-1}}=1\Leftrightarrow2^{2x-1}-1=2^{x-1}\Leftrightarrow\frac{\left(2^x\right)^2}{2}-\frac{2^x}{2}-1=0\)
Giải pt bậc hai được 2x = 2 ↔ x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
Đk:\(x\ne0;1;2;3;4\)
\(pt\Leftrightarrow\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}=2-\frac{1}{4-x}\)
\(\Leftrightarrow\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x}=2-\frac{1}{4-x}\)
\(\Leftrightarrow\frac{1}{x-4}-\frac{1}{x}=2-\frac{1}{4-x}\)\(\Leftrightarrow\frac{4}{x\left(x-4\right)}=\frac{2x-7}{x-4}\)
Dễ thấy \(x\ne4\) nên nhân 2 vế của pt vừa biến đổi với \(x-4\) ta dc:
\(\Leftrightarrow\frac{4}{x}=2x-7\Leftrightarrow x\left(2x-7\right)=4\)
\(\Leftrightarrow2x^2-7x=4\Leftrightarrow2x^2-7x-4=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x+1\right)=0\)\(\Leftrightarrow x=-\frac{1}{2}\left(x\ne4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Qui đồng rồi khử mẫu ta được:
3(3x+2)-(3x+1)=2x.6+5.2
<=> 9x+6-3x-1 = 12x+10
<=> 9x-3x-12x = 10-6+1
<=> -6x = 5
<=> x = -5/6
Vậy ....
b) ĐKXĐ: \(x\ne\pm2\)
Qui đồng rồi khử mẫu ta được:
(x+1)(x+2)+(x-1)(x-2) = 2(x2+2)
<=> x2+3x+2+x2-3x+2 = 2x2+4
<=> x2+x2-2x2+3x-3x = 4-2-2
<=> 0x = 0
<=> x vô số nghiệm
Vậy x vô số nghiệm với x khác 2 và x khác -2
c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)
\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)
Vậy ......
d) (x+1)2-4(x2-2x+1) = 0
<=> x2+2x+1-4x2+8x-4 = 0
<=> -3x2+10x-3 = 0
giải phương trình
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(A=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{2\cdot3x}{3x\left(x+1\right)}-\frac{3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right]\cdot\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\frac{x+1}{2\cdot\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{\left(-8x^2+2\right)\left(x+1\right)}{3x\left(x+1\right)2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-4x^2\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-2x\right)\left(1-2x\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{1+2x}{3x}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2x+1-3x-1+x^2}{3x}\)
\(A=\frac{x^2-x}{3x}\)
\(A=\frac{x\left(x-1\right)}{3x}\)
\(A=\frac{x-1}{3}\)
b) Thay x = 4 ta có :
\(A=\frac{4-1}{3}=\frac{3}{3}=1\)
c) Để A thuộc Z thì \(x-1⋮3\)
\(\Rightarrow x-1\in B\left(3\right)=\left\{0;3;6;...\right\}\)
\(\Rightarrow x\in\left\{1;4;7;...\right\}\)
Vậy.....
de sai ban oi