tìm x biết 3^x+3^x-1+3^x-2=1053.Mik ko chép sai đề nha đừng nhầm thành 3^x+3^x+1+3^x+2 nha.Giúp mình với ,tick liền nha nhứ giải chi tiết cụ thể giùm
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b/ (x + 5)(x + 2) - 3(4x - 3) = (5 - x)2
=> x2 + 7x + 10 - 12x + 9 = 25 - 10x + x2
=> x2 + 7x + 10 - 12x + 9 - 25 + 10x - x2 = 0
=> 5x - 6 = 0
=> x = 6/5
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt A=(x+1)+(X+2)+(x+3)+....+(x+99)
=> A= x+1+x+2+x+3+....+x+100
=x+x+x+x+...+x+(1+2+3+4+..+99)( có 99x)
=> 99x+4950=0
=> 99x=-4950
=> x=-50
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\text{*}\right)\) Tìm giá trị lớn nhất của biểu thức sau:
Ta có:
\(A=\frac{x^2+1}{x^2-x+1}=\frac{2\left(x^2-x+1\right)-\left(x^2-2x+1\right)}{x^2-x+1}=2-\frac{\left(x-1\right)^2}{x^2-x+1}\le2\) với mọi \(x\)
Dấu \("="\) xảy ra \(\Leftrightarrow\) \(\left(x-1\right)^2=0\) \(\Leftrightarrow\) \(x-1=0\) \(\Leftrightarrow\) \(x=1\)
Vậy, \(A_{max}=2\) \(\Leftrightarrow\) \(x=1\)
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\(B=\frac{3-4x}{x^2+1}=\frac{4\left(x^2+1\right)-\left(4x^2+4x+1\right)}{x^2+1}=4-\frac{\left(2x+1\right)^2}{x^2+1}\le4\) với mọi \(x\)
Dấu \("="\) xảy ra \(\Leftrightarrow\) \(\left(2x+1\right)^2=0\) \(\Leftrightarrow\) \(2x+1=0\) \(\Leftrightarrow\) \(x=-\frac{1}{2}\)
Vậy, \(B_{max}=4\) \(\Leftrightarrow\) \(x=-\frac{1}{2}\)
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\(\left(\text{*}\text{*}\right)\) Tìm giá trị nhỏ nhất của biểu thức sau:
Từ \(A=\frac{x^2+1}{x^2-x+1}\)
\(\Rightarrow\) \(3A=\frac{3x^2+3}{x^2-x+1}=\frac{\left(x^2+2x+1\right)+2\left(x^2-x+1\right)}{x^2-x+1}=\frac{\left(x+1\right)^2}{x^2-x+1}+2\ge2\) với mọi \(x\)
Vì \(3A\ge2\) nên \(A\ge\frac{2}{3}\)
Dấu \("="\) xảy ra \(\Leftrightarrow\) \(\left(x+1\right)^2=0\) \(\Leftrightarrow\) \(x+1=0\) \(\Leftrightarrow\) \(x=-1\)
Vậy, \(A_{min}=\frac{2}{3}\) \(\Leftrightarrow\) \(x=-1\)
Câu b) tự giải
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\(\Leftrightarrow\frac{1}{x^2+7x+12}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+3x+2}-\frac{1}{10}=0\)
\(\Rightarrow-\frac{x^2+5x-26}{10\left(x+1\right)\left(x+4\right)}=0\)
\(\Rightarrow x^2+5x-26=0\)
\(\Rightarrow5^2-\left(-4\left(1.26\right)\right)=129\)(cái này là D)
\(\Rightarrow x_{1,2}=\frac{-b+-\sqrt{D}}{2a}=\frac{-5+-\sqrt{129}}{2}\)
\(x=+-\frac{\sqrt{129}}{2}-2\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2001}{2003}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2003}\)
\(\Rightarrow x+1=2003\)
\(x=2002\)
Vậy x = 2002
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3+8x^2+17x+10\)
\(=x^3+2x^2+x^2+5x^2+10x+5x+2x+10\)
\(=\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(5x^2+5x\right)+\left(10x+10\right)\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+5x\left(x+1\right)+10\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+5x+10\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(2x-1\right)\left(3x-126\right)=0\Rightarrow\hept{\begin{cases}x=0\\2x-1=0\\3x-126=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0\\2x=1\\3x=126\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\frac{1}{2}\\x=42\end{cases}}\)
3x+3x-1+3x-2=1053
=> 3x-2.32+3x-2.3+3x-2=1053
3x-2.9+3x-2.3+3x-2=1053
=>3x-2.(9+3+1)=1053
3x-2.13=1053
3x-2=1053:13=81
3x-2=34
=>x-2=4
x=4+2
x=6
3*x + 3*x - 1 + 3*x - 2 = 1053
3*x + 3*x + 3*x - 3 = 1053
3*x + 3*x + 3*x = 1053 + 3 = 1056
3*x. 3 = 1056
3*x = 1056 : 3 = 352
x.x.x = 352
đấy bạn tự tính nha
Nhớ k cho mik, thank nhìu