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\(a,\Rightarrow2x^2-18x-2x^2=0\\ \Rightarrow-18x=0\Rightarrow x=0\\ b,\Rightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Rightarrow5x=22\Rightarrow x=\dfrac{22}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
câu a:
Đặt \(x-1=a\)thì pt trở thành \(\left(a+2\right)^4+\left(a-2\right)^4=82\), phá ra rồi giải pt tích
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\(x=\frac{1}{2}x+\frac{1}{4}x+\frac{1}{7}x+3\)
\(\Rightarrow x=x\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{7}\right)+3\)
\(\Rightarrow x=\frac{25}{28}x+3\Rightarrow x-\frac{25}{28}x=3\Rightarrow x\left(1-\frac{25}{28}\right)=3\Rightarrow x.\frac{3}{28}=3\Rightarrow x=28\)
Vậy x = 28
Đặt x làm nhân tử chung : x(1/2+1/4+1/7)+3=x
Qui đồng cái tổng đấy chuyển vế là ra
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\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-5}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\\ \Leftrightarrow\dfrac{5\left(x+3\right)+4\left(x-3\right)}{x^2-9}=\dfrac{x-5}{x^2-9}\\ \Leftrightarrow5x+15+4x-12=x-5\\ \Leftrightarrow5x+4x-x=-5-15+12\\ \Leftrightarrow8x=-8\\ \Leftrightarrow x=-1\left(TM\right)\\ Vậy:S=\left\{-1\right\}\)
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a: \(\Leftrightarrow x+2016=0\)
hay x=-2016
b: \(\Leftrightarrow x-100=0\)
hay x=100
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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)
a) \(\left(x-2\right)^2=\left(x-4\right)\left(x+4\right)\)
\(\Leftrightarrow x^2-4x+4-x^2+16=0\)
\(\Leftrightarrow20-4x=0\)
\(\Leftrightarrow4x=20\)
\(\Leftrightarrow x=5\)
Vậy S = {5}
b) ĐKXĐ: \(x\ne0;x\ne-2\)
\(\dfrac{x+2}{x}=\dfrac{\left(x+1\right)\left(x+4\right)}{x^2+2x}+\dfrac{x}{x+2}\)
\(\Leftrightarrow\dfrac{x+2}{x}=\dfrac{x^2+4x+x+4+x^2}{x\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{x+2}{x}=\dfrac{2x^2+5x+4}{x\left(x+2\right)}\)
\(\Rightarrow x\left(x+2\right)^2=x\left(2x^2+5x+4\right)\)
\(\Leftrightarrow x^3+4x^2+4x=2x^3+5x^2+4x\)
\(\Leftrightarrow x^3+x^2=0\)
\(\Leftrightarrow x^2\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(TM\right)\end{matrix}\right.\)
Vậy S = {-1}
c) Câu này mình không chắc về đề lắm! Bạn dùng ô chữ M bị ngược để viết lại đề nhé!
a) Ta có: \(\left(x-2\right)^2=\left(x-4\right)\left(x+4\right)\)
\(\Leftrightarrow x^2-4x+4=x^2-16\)
\(\Leftrightarrow x^2-4x+4-x^2+16=0\)
\(\Leftrightarrow-4x+20=0\)
\(\Leftrightarrow-4x=-20\)
hay x=5
Vậy: S={5}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{2x - (x - 3)(5 - x) = (x+4)}^2.\)
\(\Leftrightarrow2x-\left(5x-x^2-15+3x\right)=x^2+8x+16.\)
\(\Leftrightarrow2x-5x+x^2+15-3x-x^2-8x-16=0.\)
\(\Leftrightarrow-14x-1=0.\Leftrightarrow x=\dfrac{-1}{14}.\)
\(\text{(4x + 1)(x - 2) + 25 = (2x+3)}^2-4x.\)
\(\Leftrightarrow4x^2-8x+x-2+25=4x^2+12x+9-4x.\)
\(\Leftrightarrow-15x+14=0.\Leftrightarrow x=\dfrac{14}{15}.\)
a/ (x + 3)4 + (x + 5)4 = 16
=> (x2 + 6x + 9)2 + (x2 + 10x + 25)2 = 16
=> x4 + 36x2 + 81 + 12x3 + 108x + 18x2 + x4 + 100x2 + 625 + 20x3 + 500x + 50x2 = 16
=> 2x4 + 32x3 + 204x2 + 608x + 690 = 0
=> 2(x + 3)(x + 5)(x2 + 8x + 23) = 0
=> (x + 3)(x + 5)(x2 + 8x + 23) = 0
=> x = -3
hoặc x = -5
hoặc x2 + 8x + 23 = 0 , mà x2 + 8x + 23 > 0 => pt vô nghiệm
Vậy x = -3 , x = -5
b/ tương tự như câu a ^^