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\(-2\left(\sqrt{1+x}+\sqrt{1-x}\right)+7=\sqrt{\left(5-2x\right)\left(5+2x\right)}-2\sqrt{1-x^2}\)
ĐKCĐ: \(-1\le x\le1\)
\(\Leftrightarrow2\left(\sqrt{\left(1-x\right)}-1\right)\left(\sqrt{1+x}-1\right)+5-\sqrt{\left(5-2x\right)\left(5+2x\right)}=0\)
\(\Leftrightarrow2x^2\left[\frac{2}{5+\sqrt{\left(5-2x\right)\left(5+2x\right)}}-\frac{1}{\left(\sqrt{1-x}+1\right)\left(\sqrt{1+x}+1\right)}\right]\)
Đặt: \(A=\frac{2}{5+\sqrt{\left(5-2x\right)\left(5+2x\right)}}-\frac{1}{\left(\sqrt{1-x}+1\right)\left(\sqrt{1+x}+1\right)}\)
Có: \(A\le\frac{2}{5+\sqrt{\left(5-2\right)\left(5-2\right)}}-\frac{1}{\sqrt{1-x^2}+1+\sqrt{1-x}+\sqrt{1+x}}< \frac{2}{5+3}-\frac{1}{1+1+2}=0\)
\(\Rightarrow x=0\) là nghiệm của pt
\(x^5+y^5-\left(x+y\right)^5\)
\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+8xy^4+y^5\right)\)
\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=-5xy\left[\left(x+y\right)\left(x^2-xy+y^2\right)+2xy\left(x+y\right)\right]\)
\(=-5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)
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) \(\frac{1}{5}\left(x-\frac{1}{3}\right)-4\left(\frac{x}{5}-\frac{1}{2}\right)=x\)
\(\Leftrightarrow\frac{1}{5}x-\frac{1}{15}-\frac{4}{5}x+2=x\)
\(\Leftrightarrow\frac{1}{5}x-\frac{4}{5}x-x=\frac{1}{15}-2\)
\(\Leftrightarrow-\frac{8}{5}x=-\frac{29}{15}\)
\(\Leftrightarrow x=\frac{29}{24}\)
a, \(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
\(\Leftrightarrow x^4+2x^3+x^2+4x^2+4x+12=0\)
\(\Leftrightarrow x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^3+3x^2+8x+12\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^3+2x^2+x^2+2x+6x+12\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)=0\)
có : \(x^2+x+6>0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)
b, \(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)-297=0\)
\(\Leftrightarrow\left[\left(x-1\right)\left(x+5\right)\right]\left[\left(x-3\right)\left(x+7\right)\right]-297=0\)
\(\Leftrightarrow\left(x^2+4x-5\right)\left(x^2+7x-21\right)-297=0\)
đặt \(x^2+4x-13=t\)
\(\Leftrightarrow\left(t+8\right)\left(t-8\right)-297=0\)
\(\Leftrightarrow t^2-64-297=0\)
\(\Leftrightarrow t^2=361\)
\(\Leftrightarrow t=\pm19\)
có t rồi tìm x thôi
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 \(x\ge1\Rightarrow x=2\)
Theo bài ra , ta có :
\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)
ĐKXĐ : \(x\ne3,x\ne-3,x\ne-\frac{7}{2}\)
Quy đồng và khử mẫu phương trình ta đk :
\(13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)
\(\Leftrightarrow\left(x+3\right)\left(13+x-3\right)=6\left(2x+7\right)\)
\(\Leftrightarrow\left(x+3\right)\left(x+10\right)=12x+42\)
\(\Leftrightarrow x^2+13x+30=12x+42\)
\(\Leftrightarrow x^2+13x-12x+30-42=0\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow x^2-3x+4x-12=0\)
\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
Kết hợp với ĐKXĐ ta có : x = -4
Vậy \(S=\left\{-4\right\}\)
Chúc bạn học tốt =))
ĐKXĐ: x\(\ne\)3;-7/2;-3
\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\Leftrightarrow\frac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\frac{\left(x-3\right)\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\frac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)
\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)
\(\Leftrightarrow13x+39+x^2-9=12x+42\\ \Leftrightarrow x^2+x=12\)
\(\Leftrightarrow x^2+x-12=0\Leftrightarrow x^2-3x+4x-12=0\\ \Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\Leftrightarrow\left[\begin{matrix}x-3=0\Rightarrow x=3\\x+4=0\Rightarrow x=-4\end{matrix}\right.\)
Nhận thấy x=3 không thỏa mãn ĐKXĐ nên pt có 1 nghiệm duy nhất là x=-4
( 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