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\(a,x^4+2x^3-3x^2-8x-4=0\\ \Leftrightarrow x^4+x^3+x^3+x^2-4x^2-4x-4x-4=0\\ \Leftrightarrow x^3\left(x+1\right)+x^2\left(x+1\right)-4x\left(x+1\right)-4\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^3+x^2-4x-4\right)=0\\ \Leftrightarrow\left(x+1\right)\left[x^2\left(x+1\right)-4\left(x+1\right)\right]=0\\ \Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(x+1\right)^2\left(x-2\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=2\end{matrix}\right.\\ Vậy.....\)
\(b,\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\\ \Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\\ \Leftrightarrow\left(x^2-7+3\right)\left(x^2-7-3\right)=72\\ \Leftrightarrow\left(x^2-7\right)^2-9=72\\ \Leftrightarrow\left(x^2-7\right)^2=81\\ \Rightarrow\left[{}\begin{matrix}x^2-7=9\\x^2-7=-9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=\sqrt{-2}\left(vôlí\right)\end{matrix}\right.\\ Vậyx=\sqrt{2}\)
\(c,2x^3+7x^2+7x+2=0\\ \Leftrightarrow2x^3+2x^2+5x^2+5x+2x+2=0\\ \Leftrightarrow2x^2\left(x+1\right)+5x\left(x+1\right)+2\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(2x^2+5x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\2x^2+5x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=?\left(tựtính\right)\end{matrix}\right.\)
\(\left(3x-4\right)^2-4\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(3x-4\right)^2-\left(2x+2\right)^2=0\)
\(\Leftrightarrow\left(3x-4-2x-2\right)\left(3x-4+2x+2\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(5x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=\frac{2}{5}\end{cases}}\) ( thỏa mãn )
Vậy : ...
1/ \(\left(3x-4\right)^2-4\left(x+1\right)^2=0\)
\(\Leftrightarrow9x^2-24x+16-4\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow9x^2-24x+16-4x^2-8x-4=0\)
\(\Leftrightarrow5x^2-32x+12=0\)
\(\Leftrightarrow5x^2-30x-2x+12=0\)
\(\Leftrightarrow5x\left(x-6\right)-2\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(5x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\5x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=\frac{2}{5}\end{cases}}\)
Vậy tập nghiệm của phương trình là : \(S=\left\{6;\frac{2}{5}\right\}\)
2/ \(x^4+2x^3-3x^2-8x-4=0\)
\(\Leftrightarrow x^4+2x^3-3x^2-6x-2x-4=0\)
\(\Leftrightarrow x^3\left(x+2\right)-3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-3x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3+2x^2+x-2x^2-4x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x\left(x^2+2x+1\right)-2\left(x^2+2x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1\right)^2\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x+2=0\)
hoặc \(x+1=0\)
hoặc \(x-2=0\)
\(\Leftrightarrow\)\(x=2\)
hoặc \(x=-1\)
hoặc \(x=2\)
Vậy tập nghiệm của phương trình là \(S=\left\{2;-2;-1\right\}\)
a) Khai triển bình phương ròii giải như bình thường
b) <=>(x+2)(x2-2x+1)=0
sau đó tiếp tục giải phương trình tích là ra
c) <=>x (2x2-5x-7)=0
<=> x=0
hoặc 2x2-5x-7=0
bn đọc tự giải^^
#hoctốt
#plsss...k☺
(x2 + x + 1)(6 - 2x) = 0
<=> 6 - 2x = 0 (do x2 + x + 1 > 0)
<=> 2x = 6
<=> x = 3
Vậy S = {3}
(8x - 4)(x2 + 2x + 2) = 0
<=> 8x - 4 = 0 (vì x2 + 2x + 2 > 0)
<=> 8x = 4
<=> x = 1/2
Vậy S = {1/2}
x3 - 7x + 6 = 0
<=> x3 - x - 6x + 6 = 0
<=> x(x2 - 1) - 6(x - 1) = 0
<=> x(x - 1)(x + 1) - 6(x - 1) = 0
<=> (x2 + x - 6)(x - 1) = 0
<=> (x2 + 3x - 2x - 6)(x - 1) = 0
<=> (x + 3)(x - 2)(x - 1) = 0
<=> x + 3 = 0
hoặc x - 2 = 0
hoặc x - 1 = 0
<=> x = -3
hoặc x = 2
hoặc x = 1
Vậy S = {-3; 1; 2}
x5 - 5x3 + 4x = 0
<=> x(x4 - 5x2 + 4) = 0
<=> x(x4 - x2 - 4x2 + 4) = 0
<=> x[x2(x2 - 1) - 4(x2 - 1)] = 0
<=> x(x - 2)(x + 2)(x - 1)(x + 1) = 0
<=> x = 0 hoặc x - 2 = 0 hoặc x + 2 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0
<=> x = 0 hoặc x = 2 hoặc x = -2 hoặc x = 1 hoặc x = -1
Vậy S = {-2; -1; 0; 1; 2}
+ Ta có: \(\left(x^2+x+1\right).\left(6-2x\right)=0\)
- Ta lại có: \(x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
- Vì \(x^2+x+1>0\forall x\)mà \(\left(x^2+x+1\right).\left(6-2x\right)=0\)
\(\Rightarrow6-2x=0\Leftrightarrow-2x=-6\Leftrightarrow x=3\left(TM\right)\)
Vậy \(S=\left\{3\right\}\)
+ Ta có: \(\left(8x-4\right).\left(x^2+2x+2\right)=0\)
- Ta lại có: \(x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1>0\forall x\)
- Vì \(x^2+2x+2>0\forall x\)mà \(\left(8x-4\right).\left(x^2+2x+2\right)=0\)
\(\Rightarrow8x-4=0\Leftrightarrow8x=4\Leftrightarrow x=\frac{1}{2}\left(TM\right)\)
Vậy \(S=\left\{\frac{1}{2}\right\}\)
+ Ta có: \(x^3-7x+6=0\)
\(\Leftrightarrow\left(x^3-x^2\right)+\left(x^2-x\right)+\left(6x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left[\left(x^2-2x\right)+\left(3x-6\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right).\left(x-2\right).\left(x+3\right)=0\)
Vậy \(S=\left\{-3;1;2\right\}\)
+ Ta có: \(x^5-5x^3+4x=0\)
\(\Leftrightarrow x.\left[\left(x^4-x^2\right)-\left(4x^2-4\right)\right]=0\)
\(\Leftrightarrow x.\left[x^2.\left(x^2-1\right)-4.\left(x^2-1\right)\right]=0\)
\(\Leftrightarrow x.\left(x^2-1\right).\left(x^2-4\right)=0\)
\(\Leftrightarrow x=0\left(TM\right)\)
hoặc \(x^2-1=0\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\left(TM\right)\)
hoặc \(x^2-4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\left(TM\right)\)
Vậy \(S=\left\{-2;-1;0;1;2\right\}\)
!!@@# ^_^ Chúc bạn hok tốt ^_^#@@!!
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
a) đặt \(\left(x^2+x\right)\)là \(y\)
ta có: \(3y^2-7y+4\)\(=0\)
<=>\(\left(3y-4\right)\left(y-1\right)=0\)
còn lại bạn tự xử nhé
\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)
<=>\(\left(x^2-4\right)\left(x^2-10\right)=72\) (1)
Đặt \(x^2-7=t\)
=> pt (1) <=> \(\left(t+3\right)\left(t-3\right)=72\)
<=> \(t^2-9=72\)
<=> \(t^2-81=0\)
<=> \(\left(t-9\right)\left(t+9\right)=0\)
Tự làm nốt
\(8x^2-\left(4x+3\right)^3+\left(2x+3\right)^3=0\)
\(\Leftrightarrow8x^2+\left(2x+3-4x-3\right)\left[\left(4x+3\right)^2+\left(2x+3\right)\left(4x+3\right)+\left(2x+3\right)^2\right]=0\)
\(\Leftrightarrow8x^2-2x\left(16x^2+24x+9+8x^2+18x+9+4x^2+12x+9\right)=0\)
\(\Leftrightarrow2x\left(4x-28x^2-54x-27\right)=0\)
\(\Leftrightarrow2x\left(28x^2+50x+27\right)=0\)
Tự làm nốt
a) \(x^4+2x^3-3x^2-8x-4=0\)
\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+5x^2-10x+2x-4=0\)
\(\Leftrightarrow x^3\left(x-2\right)+4x^2\left(x-2\right)+5x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+4x^2+5x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+x^2+3x^2+3x+2x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+1\right)+3x\left(x+1\right)+2\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2+3x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2+2x+x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left[x\left(x+2\right)+\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)^2\left(x+2\right)=0\)
\(\Rightarrow x\in\left\{2;-1;-2\right\}\)
Vậy....
c, \(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow2\left(x^3+1\right)+7x\left(x+1\right)=0\Leftrightarrow2\left(x+1\right)\left(x^2-x+1\right)+7x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[2\left(x^2-x+1\right)+7x\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2+5x+2\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(2x+1\right)=0\)
Tập nghiệm của pt: \(S=\left\{-1;-2;-\frac{1}{2}\right\}\)
b, \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\) (1)
Đặt: \(x^2-7=t\left(t\ge-7\right)\)
Khi đó (1) trở thành: \(\left(t+3\right)\left(t-3\right)=72\Leftrightarrow t^2-9=72\Leftrightarrow\orbr{\begin{cases}t=9\\t=-9\left(loai\right)\end{cases}}\)
\(t=9\Rightarrow x^2-7=9\Leftrightarrow x=\pm4\)
Tập nghiệm của pt là \(S=\left\{\pm4\right\}\)
a, \(x^4+2x^3-3x^2-8x-4=0\)
\(\Leftrightarrow x^3\left(x+1\right)+x^2\left(x+1\right)-4x\left(x+1\right)-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+x^2-4x-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm2\end{cases}}\)