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Đặt √(x+1) làm thừa số chung rồi phân tích tiếp. Nghiệm là 0 và 3
\(x^5+y^5-\left(x+y\right)^5\)
\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+8xy^4+y^5\right)\)
\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=-5xy\left[\left(x+y\right)\left(x^2-xy+y^2\right)+2xy\left(x+y\right)\right]\)
\(=-5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)
\(\frac{2x}{x+1}=\frac{x^2-x+8}{\left(x+1\right)\left(x+1\right)}\) (ĐKXĐ: x \(\ne\) 1)
\(\Leftrightarrow\) \(\frac{2x\left(x+1\right)}{\left(x+1\right)\left(x+1\right)}=\frac{x^2-x+8}{\left(x+1\right)\left(x+1\right)}\)
\(\Leftrightarrow\) 2x(x + 1) = x2 - x + 8
\(\Leftrightarrow\) 2x2 + 2x - x2 + x - 8 = 0
\(\Leftrightarrow\) x2 + 3x - 8 = 0
\(\Leftrightarrow\) (x + \(\frac{3}{2}\))2 - \(\frac{41}{4}\) = 0
\(\Leftrightarrow\) (x + \(\frac{3}{2}\) - \(\frac{\sqrt{41}}{2}\))(x + \(\frac{3}{2}\) + \(\frac{\sqrt{41}}{2}\)) = 0
\(\Leftrightarrow\) (x + \(\frac{3-\sqrt{41}}{2}\))(x + \(\frac{3+\sqrt{41}}{2}\)) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3-\sqrt{41}}{2}=0\\x+\frac{3+\sqrt{41}}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{41}-3}{2}\\x=\frac{-3-\sqrt{41}}{2}\end{matrix}\right.\)
Vậy S = {\(\frac{\sqrt{41}-3}{2}\); \(\frac{-3-\sqrt{41}}{2}\)}
Chúc bn học tốt!!
\(\frac{2x}{x+1}=\frac{x^2-x+8}{\left(x+1\right)\cdot\left(x-4\right)}\\ \Leftrightarrow\frac{2x^2-8x}{\left(x+1\right)\cdot\left(x-4\right)}-\frac{x^2-x+8}{\left(x+1\right)\cdot\left(x-4\right)}=0\\ \Leftrightarrow\frac{2x^2-8x-x^2+x-8}{\left(x+1\right)\cdot\left(x-4\right)}=0\\ \Leftrightarrow\frac{x^2-7x-8}{\left(x+1\right)\cdot\left(x-4\right)}=0\\ \Leftrightarrow x^2-7x-8=0\\ \Rightarrow\left[{}\begin{matrix}x=\frac{7+\sqrt{17}}{2}\\x=\frac{7-\sqrt{17}}{2}\end{matrix}\right.\)
a) \(2\left(3x-1\right)-\left(5+3x\right)=3\left(2x-1\right)\)
\(\Leftrightarrow6x-2-5-3x=6x-3\)
\(\Leftrightarrow6x-3x-6x=-3+2+5\)
\(\Leftrightarrow-3x=4\)
\(\Leftrightarrow x=-\frac{4}{3}\)
b) \(3\left(x-\frac{1}{2}\right)+4\left(\frac{x}{3}-\frac{1}{3}\right)=\frac{x}{4}\)
\(\Leftrightarrow3x-\frac{3}{2}+\frac{4}{3}x-\frac{4}{3}=\frac{x}{4}\)
\(\Leftrightarrow3x+\frac{4}{3}x-\frac{x}{4}=\frac{3}{2}+\frac{4}{3}\)
\(\Leftrightarrow\frac{49}{12}x=\frac{17}{6}\)
\(\Leftrightarrow x=\frac{34}{49}\)
c) \(\frac{1}{5}\left(x-\frac{1}{3}\right)-4\left(\frac{x}{5}-\frac{1}{2}\right)=x\)
\(\Leftrightarrow\frac{1}{5}x-\frac{1}{15}-\frac{4}{5}x+2=x\)
\(\Leftrightarrow\frac{1}{5}x-\frac{4}{5}x-x=\frac{1}{15}-2\)
\(\Leftrightarrow-\frac{8}{5}x=-\frac{29}{15}\)
\(\Leftrightarrow x=\frac{29}{24}\)
( x - 1 ) ( x - 3 ) ( x + 5 ) ( x + 7) - 297 = 0
<=> ( x2 + 4x - 5 ) ( x2 + 4x - 21 ) - 297 = 0
Đặt x2 + 4x - 5 = t ( t > -9 )
Ta có : t (t - 16 ) - 297 = 0 <=> t2 - 16t - 297 = 0 <=> t = 27 ; t = 11 ( loại)
Ta có x2 + 4x - 5 = 27 <=> x2 + 4x - 32 = 0 <=> x = 4 , x = -8
Câu hỏi của Phương Boice - Toán lớp 8 - Học toán với OnlineMath
Đặt \(\sqrt{x^2-x+1}=a\left(ĐK:a>0\right)\)
\(pt\Leftrightarrow\frac{\left(x^6+3x^4a\right)\left(4-a^2\right)}{4\left(2+a\right)a^2}=a\left(2-a\right)\)
\(\Leftrightarrow\left(x^6+3x^4a\right)\left(4-a^2\right)=4a^3\left(4-a^2\right)\)
\(\Leftrightarrow\left(4-a^2\right)\left(x^6+3x^4a-4a^3\right)=0\)
TH1: \(4-a^2=0\Leftrightarrow\orbr{\begin{cases}a=-2\left(l\right)\\a=2\left(n\right)\end{cases}}\)
Với a = 2 , \(\sqrt{x^2-x+1}=2\Rightarrow x^2-x-3=0\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}+1}{2}\\x=\frac{-\sqrt{13}+1}{2}\end{cases}}\)
TH2: \(x^6+3x^4a-4a^3=0\Rightarrow x^6-x^4a+4x^4a-4x^2a^2+4x^2a^2-4a^3=0\)
\(\Leftrightarrow\left(x^2-a\right)\left(x^4+4x^2a+4a^2\right)=0\Leftrightarrow\left(x^2-a\right)\left(x^2+2a\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=a\\x^2=-2a\left(l\right)\end{cases}}\)
Với \(x^2=a\Rightarrow x^2=\sqrt{x^2-x+1}\)
Đến đây bình phương và tìm ra nghiệm.
Ta có:\(\left|x+1\right|+\left|x^2+x-2\right|\ge0\)
\(\Rightarrow x^3-1\ge0\)
\(\Leftrightarrow x\ge1\)
\(\Rightarrow pt\Leftrightarrow x+1+x^2+x-2=x^3-1\)
\(\Leftrightarrow x^2+2x-1=x^3-1\)
\(\Leftrightarrow x\left(x^2-x-2\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-2\right)=0\)
Mà \(x\ge1\Rightarrow x=2\)