Tính giá trị biểu thức: 2+4+6+...+2x=210
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![](https://rs.olm.vn/images/avt/0.png?1311)
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Ta có : 2 + 4 + 6 +......+ 2x = 210
=> \(\frac{\left[\left(2x-2\right):2+1\right]\left(2x+2\right)}{2}=210\)
=> \(\frac{\left[x-1+1\right]\left(2x+2\right)}{2}=210\)
=> \(\frac{2x\left(x+1\right)}{2}=210\)
=> x(x + 1) = 210
=> x (x + 1) = 14.15
=> x = 14
Vậy x = 14 .
![](https://rs.olm.vn/images/avt/0.png?1311)
Q=x^6+x^5+x^5+x^4+x^4+x^3+x^3+x^2+x^2+x+x+1
=x^4(x^2+x)+x^3(x^2+x)+x^2(x^2+x)+x(x^2+x)+1+x+1
=x^4+x^3+x^2+x+x+2
=x^4+x^3+x^2+2x+2
=x^2(x^2+x)+x^2+x+x+2
=x^2+1+x+2
=x^2+x+3
=1+3
=4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P\left(x\right)=x^4+ax^3+bx^2+cx+d\)
Theo đề ta có:
\(\hept{\begin{cases}1+a+b+c+d=0\\81+27a+9b+3c+d=0\\625+125a+25b+5c+d=0\end{cases}\Leftrightarrow\hept{\begin{cases}a+b+c+d=-1\\27a+9b+3c+d=-81\\125a+25b+5c+d=-625\end{cases}}\Leftrightarrow\hept{\begin{cases}a=-9\\b=23\\c=-15;\end{cases}d=-1}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Answer:
a) \(\frac{5x}{2x+2}+1=\frac{6}{x+1}\)
\(\Rightarrow\frac{5x}{2\left(x+1\right)}+\frac{2\left(x+1\right)}{2\left(x+1\right)}=\frac{12}{2\left(x+1\right)}\)
\(\Rightarrow5x+2x+2-12=0\)
\(\Rightarrow7x-10=0\)
\(\Rightarrow x=\frac{10}{7}\)
b) \(\frac{x^2-6}{x}=x+\frac{3}{2}\left(ĐK:x\ne0\right)\)
\(\Rightarrow x^2-6=x^2+\frac{3}{2}x\)
\(\Rightarrow\frac{3}{2}x=-6\)
\(\Rightarrow x=-4\)
c) \(\frac{3x-2}{4}\ge\frac{3x+3}{6}\)
\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\ge0\)
\(\Rightarrow9x-6-6x-6\ge0\)
\(\Rightarrow3x-12\ge0\)
\(\Rightarrow x\ge4\)
d) \(\left(x+1\right)^2< \left(x-1\right)^2\)
\(\Rightarrow x^2+2x+1< x^2-2x+1\)
\(\Rightarrow4x< 0\)
\(\Rightarrow x< 0\)
e) \(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}\le\frac{x^2}{7}-\frac{2x-3}{5}\)
\(\Rightarrow\frac{2x-3+5\left(x^2-2x\right)}{35}\le\frac{5x^2-7\left(2x-3\right)}{35}\)
\(\Rightarrow2x-3+5x^2-10x\le5x^2-14x+21\)
\(\Rightarrow6x\le24\)
\(\Rightarrow x\le4\)
f) \(\frac{3x-2}{4}\le\frac{3x+3}{6}\)
\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\le0\)
\(\Rightarrow9x-6-6x-6\le0\)
\(\Rightarrow3x\le12\)
\(\Rightarrow x\le4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-x=6\)
\(\Rightarrow x.\left(x^2-1\right)=6\)
\(\Rightarrow x.\left(x-1\right).\left(x+1\right)=6\)
\(x^6-2x^4+x^3+x^2-x\)
\(=x^6-x^5+x^5-x^4-x^4+x^3+x^2-x\)
\(=x^5.\left(x-1\right)+x^4.\left(x-1\right)-x^3.\left(x-1\right)+x.\left(x-1\right)\)
\(=\left(x-1\right).\left(x^5+x^4-x^3+x\right)\)
\(=\left(x-1\right).[x^4.\left(x+1\right)-x.\left(x^2-1\right)]\)
\(=\left(x-1\right).\left(x+1\right).[x^4-x.\left(x-1\right)]\)
\(=\left(x-1\right).\left(x+1\right).\left(x^4-x^2+x\right)\)
\(=x.\left(x-1\right).\left(x+1\right).\left(x^3-x+1\right)\)
\(=6.\left(6+1\right)\)
\(=42\)
Vậy giá trị của biểu thức \(B=42\)khi \(x^3-x=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
2+4+6+....+2x=210
\(\Rightarrow\frac{\left(2x+2\right).\left[\left(2x-2\right):2+1\right]}{2}=210\)
\(\Rightarrow\left(2x+2\right).x=210.2=420\)
\(\Rightarrow2.x^2+2.x=420\)
\(\Rightarrow2\left(x^2+x\right)=420\)
\(\Rightarrow x^2+x=210\)
\(\Rightarrow x\left(x+1\right)=210=14.15\)
\(\Rightarrow x=14\)
Tổng các số chẵn từ 2 đến 8 là : (2+8).[(8-2):2+1]:2=20
Tổng các số còn lại là : 210 - 20 = 190
Tổng các số chẵn từ 10 đến 18 là : (18+10).[(18-10):2+1):2=70
Tổng các số chẵn từ 20 đến 28 là : 190 - 70 = 20
Vậy x = 8.