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a) (x-1)(5x+3)=(3x-8)(x-1)
= (x-1)(5x+3)-(3x-8)(x-1)=0
=(x-1)[(5x+3)-(3x-8)]=0
=(x-1)(5x+3-3x+8)=0
=(x-1)(2x+11)=0
\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0
\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)
Vậy S={1;\(\dfrac{-11}{2}\)}
b) 3x(25x+15)-35(5x+3)=0
=3x.5(5x+3)-35(5x+3)=0
=15x(5x+3)-35(5x+3)=0
=(5x+3)(15x-35)=0
\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0
\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)
Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}
c) (2-3x)(x+11)=(3x-2)(2-5x)
=(2-3x)(x+11)-(3x-2)(2-5x)=0
=(3x-2)[(x+11)-(2-5x)]=0
=(3x-2)(x+11-2+5x)=0
=(3x-2)(6x+9)=0
\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0
\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)
Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}
d) (2x2+1)(4x-3)=(2x2+1)(x-12)
=(2x2+1)(4x-3)-(2x2+1)(x-12)=0
=(2x2+1)[(4x-3)-(x-12)=0
=(2x2+1)(4x-3-x+12)=0
=(2x2+1)(3x+9)=0
\(\Leftrightarrow\)2x2+1=0 hoặc 3x+9=0
\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3
Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}
e) (2x-1)2+(2-x)(2x-1)=0
=(2x-1)[(2x-1)+(2-x)=0
=(2x-1)(2x-1+2-x)=0
=(2x-1)(x+1)=0
\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0
\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1
Vậy S={\(\dfrac{-1}{2}\);-1}
f)(x+2)(3-4x)=x2+4x+4
=(x+2)(3-4x)=(x+2)2
=(x+2)(3-4x)-(x+2)2=0
=(x+2)[(3-4x)-(x+2)]=0
=(x+2)(3-4x-x-2)=0
=(x+2)(-5x+1)=0
\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0
\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)
Vậy S={-2;\(\dfrac{1}{5}\)}
(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 ^_^#@@!!
ko ai giải đc à, giúp mk đi mà mau lên đang cần gấp, please
RẤT nhieu bn giai dc vi các pt này dễ nhung k ai giai vi nó dài ,làm mệt mà kè nhờ vả k biet ơn, k coi trọng chât xám
toàn là h tảo lao nên ng tài k dc trọng dụng , kẻ bât tai thi k giai dc, bởi z ng tài chỉ xem bài nào khó, k dài thi giai, dc kdc h cũng k cần
a)\(\left(x^2+1\right)\left(x^2-4x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-1\left(vn\right)\\\left(x-2\right)^2=0\end{cases}\Rightarrow}x=2}\)
b)\(\left(3x-2\right)\left(\frac{2x+6}{7}-\frac{4x-3}{5}\right)=0\\ \Rightarrow\left(3x-2\right)\left(\frac{10x+30-28x+21}{35}\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(\frac{-18x+51}{35}\right)=0\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}\)
c)\(\left(3,3-11x\right)\left(\frac{21x+6+10-30x}{15}\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{16}{9}\end{cases}}\)
câu a, b, c dễ mà. Bạn áp dụng 7 hằng đẳng thúc là làm đc thoii!!
vd: a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=\left(3x+2\right)\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)[\left(3x-2\right)-\left(x-1\right)]=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\) (bạn phá ngoặc ra rồi tính là ra bước này)
\(\Leftrightarrow3x+2=0\) hoặc \(x+1=0\) hoặc \(2x-1=0\) ( đến đây bạn chia làm 3 trường hợp r tự tính nhé)
Chúc bạn học tốt!!
d/
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
e/
\(\Leftrightarrow x^3+x^2-6x-x^2-x+6=0\)
\(\Leftrightarrow x\left(x^2+x-6\right)-\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
TA CÓ:
\(a,\left(4x-1\right)\left(x-3\right)=\left(x-3\right)\left(5x+2\right)\Leftrightarrow\left(4x-1\right)\left(x-3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\left(x-3\right)\left(4x-1-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-x-3\right)=0\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
\(b,\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\Leftrightarrow\left(x+3\right)\left(x-5+3x-4\right)=0\)
\(\left(x-3\right)\left(4x-9\right)=0\orbr{\begin{cases}x=3\\x=\frac{9}{4}\end{cases}}\)
\(c,\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\Leftrightarrow\left(1-x\right)\left(5x+3\right)=\left(7-3x\right)\left(1-x\right)\)
\(\left(1-x\right)\left(5x+3-7+3x\right)=0\Leftrightarrow\left(1-x\right)\left(8x-4\right)=0\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(\left(x^2+7x+12\right).\left(4x-16\right)-\left(x+3\right)\left(x^2-5x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^2+3x+4x+12\right).4.\left(x-4\right)-\left(x+3\right)\left(x^2-x-4x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow4\left(x+4\right)\left(x+3\right)\left(x-4\right)-\left(x+3\right)\left(x-4\right)\left(x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-4\right)\left(x+3\right)\left(4-x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-4\right)\left(x+3\right)\left(8-x\right)=0\)
\(\Leftrightarrow\frac{\orbr{\begin{cases}x+4=0\\x-4=0\end{cases}}}{\orbr{\begin{cases}x+3=0\\8-x=0\end{cases}}}\Leftrightarrow\frac{\orbr{\begin{cases}x=-4\\x=4\end{cases}}}{\orbr{\begin{cases}x=-3\\x=8\end{cases}}}\)