( m : 1 - m : 1 ) x 91 x m =?
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Chọn D
+ Xét bpt : 3x-4> x+ 9 hay x> 5/ 2
Suy ra tập nghiệm của bpt đầu là : S1= ( 5/2; + ∞)
+ Xét bpt: 1-2x ≤ m-3x+ 1
Hay x ≤ m
Suy ra tập nghiệm của bpt thứ 2 là S2= ( -∞; m]
Để hệ bpt vô nghiệm khi và chỉ khi :
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Delta=49-4\left(m-3\right)=61-4m\ge0\Rightarrow m\le\frac{61}{4}\)
Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=7\\x_1x_2=m-3\end{matrix}\right.\)
\(x_1^3+x_2^3=91\)
\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=91\)
\(\Leftrightarrow343-21\left(m-3\right)=91\)
\(\Rightarrow m=15\) (thỏa mãn)
![](https://rs.olm.vn/images/avt/0.png?1311)
x + x + x + 91 = -2
x + x + x = -2 - 91
x + x + x = -93
3x = -93
x = -93 : 3
x = -31
-152 - (3x + 1) = (-2) . (-27)
-152 - (3x + 1) = 54
3x + 1 = -152 - 54
3x + 1 = -206
3x = -206 - 1
3x = -207
x = -207 : 3
x = -69
|5x + 1| = 11
5x + 1 = 11 hay 5x + 1 = -11
5x = 11 - 1 hay 5x = -11 - 1
5x = 10 hay 5x = -12
x = 10 : 5 hay x = -12 : 5
x = 2 hay x = -2,4
Mà x ∈ Z
x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
c)\(4x^4-101x^2+25=0\)
\(\Leftrightarrow4x^4-100x^2-x^2+25=0\)
\(\Leftrightarrow4x^2\left(x^2-25\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(4x^2-1\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+1=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\\x=5\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};\frac{-1}{2};5;-5\right\}\)
a)\(\left(2x-1\right)^2=x+5\)
\(\Leftrightarrow4x^2-4x+1=x+5\)
\(\Leftrightarrow4x^2-5x-4=0\)
\(\Leftrightarrow4\left(x^2-\frac{5}{4}x-1\right)=0\)
\(\Leftrightarrow4\left(x^2-\frac{5}{4}x+\frac{25}{64}-\frac{89}{64}\right)=0\)
\(\Leftrightarrow4\left[\left(x-\frac{5}{8}\right)^2-\frac{89}{64}\right]=0\)
\(\Leftrightarrow4\left(x-\frac{5}{8}+\frac{\sqrt{89}}{8}\right)\left(x-\frac{5}{8}-\frac{\sqrt{89}}{8}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{5}{8}+\frac{\sqrt{89}}{8}=0\\x-\frac{5}{8}-\frac{\sqrt{89}}{8}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5-\sqrt{89}}{8}\\x=\frac{5+\sqrt{89}}{8}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5-\sqrt{89}}{8};\frac{5+\sqrt{89}}{8}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63\cdot1,2-21\cdot3,6\right)}{1-2+3-4+5-6+...+99-100}\)
đề là vậy nhé mn
để ý chút thấy liền ah : 63.1,2-21.3,6=63.1,2-21.3.1,2= 63.1,2- 63.1,2=0
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Ta có P = \(\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+5-...+99-100}\)= \(\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)0}{1-2+3-4+5-...+99-100}\)= \(\dfrac{0}{1-2+3-4+5-6+...+99-100}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x\left(1+5^2\right)=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow5^x=5^2\)
\(\Leftrightarrow x=2\)
Vậy..
b/ \(2^{91}=\left(2^{13}\right)^7=8192^7\left(1\right)\)
\(5^{35}=\left(5^5\right)^7=3125^7\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow2^{91}>5^{35}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
= [ m-m] x 91 x m
= 0 x 91 x m
=0 x m
=0
= (m-m) x 19 x m
= 0 x 19 x m
= 0