Giải phương trình:
\(3x\times\left(1-x\right)+\left(x+3\right)\times\left(x-2\right)=-2\times\left(x-4\right)^2\)
Giải phương trình \(\left(x-2\right)\times\left(x+2\right)+4\left(x-2\right)\times\sqrt{\frac{x+2}{x-2}}=-3\)
ĐK \(\orbr{\begin{cases}x>2\\x\le-2\end{cases}}\)
Đặt \(\sqrt{\frac{x+2}{x-2}}=t\Rightarrow x+2=t^2\left(x-2\right)\)
Vậy thì phương trình trở thành \(t^2\left(x-2\right)^2+4\left(x-2\right)t+3=0\)
\(\Leftrightarrow\left[t\left(x-2\right)+1\right]\left[t\left(x-2\right)+3\right]=0\)
Với \(t\left(x-2\right)+1=0\Leftrightarrow\sqrt{\frac{x+2}{x-2}}\left(x-2\right)+1=0\)
Để pt có nghiệm thì \(x-2< 0\) , khi đó \(-\sqrt{\frac{x+2}{x-2}\left(x-2\right)^2}+1=0\Leftrightarrow-\sqrt{x^2-4}+1=0\)
\(\Leftrightarrow x^2-4=1\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5}\left(l\right)\\x=-\sqrt{5}\left(n\right)\end{cases}}\)
Với \(t\left(x-2\right)+3=0\Leftrightarrow-\sqrt{x^2-4}+3=0\)
\(\Leftrightarrow x^2-4=9\Leftrightarrow\orbr{\begin{cases}x=\sqrt{13}\left(l\right)\\x=-\sqrt{13}\left(n\right)\end{cases}}\)
Vậy pt có tập nghiệm \(S=\left\{-\sqrt{13};-\sqrt{5}\right\}\)
\(3x\times\left(1-x\right)+\left(x+3\right)\times\left(x-2\right)=-2\times\left(x-4\right)^2\)
Lời giải:
$3x(1-x)+(x+3)(x-2)=-2(x-4)^2$
$\Leftrightarrow (3x-3x^2)+(x^2-2x+3x-6)=-2(x^2-8x+16)$
$\Leftrightarrow -2x^2+4x-6=-2x^2+16x-32$
$\Leftrightarrow 12x=26\Rightarrow x=\frac{13}{6}$
Vậy........
Giải phương trình:
\(\left(x^2+6x+10\right)^2+\left(x+3\right)\left(3x^2+20x+36\right)=0\)0
\(\left(x^2+x\right)^2+4\times\left(x^2+x\right)=12\)
\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
Đặt \(a=x^2+x\)
\(\Leftrightarrow a^2+4a=12\)
\(\Leftrightarrow a^2+4a-12=0\)
\(\Leftrightarrow a^2+6a-2a-12=0\)
\(\Leftrightarrow a\left(a+6\right)-2\left(a+6\right)=0\)
\(\Leftrightarrow\left(a+6\right)\left(a-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-6\\a=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+x=-6\\x^2+x=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{23}{4}=0\\x^2+2x-x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\\\left(x+2\right)\left(x-1\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)
Vậy....
\(\left(x^2+6x+10\right)^2+\left(x+3\right)\left(3x^2+20x+36\right)=0\)
( rút gọn phá ngoặc tất cả )
\(\Leftrightarrow x^4+15x^3+85x^2+216x+208=0\)
\(\Leftrightarrow x^4+4x^3+11x^3+44x^2+41x^2+164x+52x+208=0\)
\(\Leftrightarrow x^3\left(x+4\right)+11x^2\left(x+4\right)+41x\left(x+4\right)+52\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^3+11x^2+41x+52\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^3+4x^2+7x^2+28x+13x+52\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left[x^2\left(x+4\right)+7x\left(x+4\right)+13\left(x+4\right)\right]=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+4\right)\left(x^2+7x+13\right)=0\)
\(\Leftrightarrow\left(x+4\right)^2\left(x^2+2\cdot x\cdot\frac{7}{2}+\frac{49}{4}+\frac{3}{4}\right)=0\)
\(\Leftrightarrow\left(x+4\right)^2\left[\left(x+\frac{7}{2}\right)^2+\frac{3}{4}\right]=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy....
Giải phương trình:\(\left(x+2\right)\times\left(x-2\right)\times\left(x^2-10\right)=72\)
Bài làm:
Ta có: \(\left(x+2\right)\left(x-2\right)\left(x^2-10\right)=72\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)
\(\Leftrightarrow x^4-14x^2+40-72=0\)
\(\Leftrightarrow x^4-14x^2-32=0\)
\(\Leftrightarrow\left(x^4-16x^2\right)+\left(2x^2-32\right)=0\)
\(\Leftrightarrow x^2\left(x^2-16\right)+2\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(x^2-16\right)=0\)
Mà \(x^2+2\ge2>0\left(\forall x\right)\)
\(\Rightarrow x^2-16=0\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow x=\pm4\)
( x + 2 )( x - 2 )( x2 - 10 ) = 72
<=> ( x2 - 4 )( x2 - 10 ) = 72
<=> x4 - 14x2 + 40 - 72 = 0
<=> x4 - 14x2 - 32 = 0
Đặt t = x2 ( \(t\ge0\))
Pt <=> t2 - 14t - 32 = 0
<=> t2 + 2t - 16t - 32 = 0
<=> t( t + 2 ) - 16( t + 2 ) = 0
<=> ( t - 16 )( t + 2 ) = 0
<=> \(\orbr{\begin{cases}t-16=0\\t+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}t=16\\t=-2\end{cases}}\)
\(t\ge0\Rightarrow t=16\)
=> x2 = 16
=> \(x=\pm4\)
1.Rút gọn biểu thức:
\(a,\)\(x\times\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(b,\)\(3x\times\left(x-2\right)-5x\times\left(1-x\right)-8\times\left(x^2-3\right)\)
\(c,\)\(\left(2x-6\right)\times\left(x+3\right)-5\times\left(2x^2-x+7\right)\)
Rút gọn \(B=\left(x^4-x+\frac{x-3}{x^3+1}\times\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right)\times\frac{4x^2+6x+1}{\left(x+3\right)\left(4-x\right)}\)
Tìm x biết:
\(\frac{-1}{2}\times\left(3x-1\right)+\frac{3}{4}\left(3-2x\right)=-3\times\left(\frac{x}{2}-1\right)-\left(\frac{4}{5}\right)^{-1}\)
Giải phương trình sau:
\(3x\times\left|x+1\right|-2x\times\left|x+2\right|=2\) với x>-1
Tìm x biết
a)\(\dfrac{2}{\left(x+2\right)\times\left(x+4\right)}+\dfrac{4}{\left(x+4\right)\times\left(x+8\right)}+\dfrac{6}{\left(x+8\right)\times\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\times\left(x+14\right)}\)
Lời giải:
PT \(\Leftrightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}\)
\(\Leftrightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}\)
\(\Leftrightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}\)
\(\Leftrightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}\)
\(\Rightarrow x=12\) (thỏa mãn)
Vậy......