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2)
a) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy x=0 ; x=-1 ; x=1
b) \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
1)
a) \(\left(x-2\right)\left(x^2+3x+4\right)\)
\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)
\(\Leftrightarrow x^3+x^2-2x-8\)
b) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
c) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
d) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)
\(=17x^2+5x-6-6x^3\)
\(a.2\left(x+3\right)-x\left(x+3\right)=0\)
\(\text{⇔}\left(x+3\right)\left(2-x\right)=0\)
\(\text{⇔}x=-3orx=2\)
\(b.x^2\left(2x+3\right)-8x-12=0\)
\(\text{⇔}x^2\left(2x+3\right)-4\left(2x+3\right)=0\)
\(\text{⇔}\left(x-2\right)\left(x+2\right)\left(2x+3\right)=0\)
\(\text{⇔}x=2;x=-2orx=-\dfrac{3}{2}\)
\(c.\left(2x-7\right)^2-\left(x-3\right)^2=0\)
\(\text{⇔}\left(2x-7-x+3\right)\left(2x-7+x-3\right)=0\)
\(\text{⇔}\left(x-4\right)\left(3x-10\right)=0\)
\(\text{⇔}x=4orx=\dfrac{10}{3}\)
\(d.\left(5x^2+3x-2\right)^2=\left(4x^2-3x-2\right)^2\)
\(\text{⇔}\left(5x^2+3x-2-4x^2+3x+2\right)\left(5x^2+3x-2+4x^2-3x-2\right)=0\)
\(\text{⇔}\left(x^2+6x\right)\left(9x^2-4\right)=0\)
\(\text{⇔}x\left(x+6\right)\left(9x^2-4\right)=0\)
\(\text{⇔}x=0;x=-6orx=+-\dfrac{2}{3}\)
Còn lại tượng tự nha , dài quá ~
a) 2(x+3)=x(x+3)
2x+6=x^2+3x
2x-x^2-3x=6
x(2-x-3)=6
x(-1-x)=6 ( xong lập bang nhà)
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
a)\(x^2+3x+6=x^2+2.\frac{3}{2}x+\frac{9}{4}+\frac{15}{4}=0\)
\(\left(x+\frac{3}{2}\right)^2+\frac{15}{4}=0\)
\(\left(x+\frac{3}{2}\right)^2=-\frac{15}{4}\)
Vì bình phương luôn lớn hơn hoặc bằng 0
Nên PT vô nghiệm
b)\(x^2-2x-3=0\)
\(x^2-3x+x-3=0\)
\(\left(x+1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
d)\(x^3-2x^2-x+2=0\)
\(x^2\left(x-2\right)-\left(x-2\right)=0\)
\(\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
x - 2 = 0 x=2
c)\(2x^2+7x+3=0\)
\(2x^2+x+6x+3=0\)
\(x\left(2x+1\right)+3\left(2x+1\right)=0\)
\(\left(2x+1\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\x+3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=-3\end{cases}}\)
a) \(\left(y-1\right)^2=9\)
\(\Rightarrow\left(y-1\right)^2=3^2=\left(-3\right)^2\)
\(\Rightarrow x-1=3\Rightarrow x=4\)
\(\Rightarrow x-1=-3\Rightarrow x=-2\)
Vậy: \(x=4\) hoặc \(-2\)
a) (5x - 1)(2x + 1) = (5x -1)(x + 3)
<=> (5x - 1)(2x + 1) - (5x -1)(x + 3) = 0
<=> (5x - 1)(2x + 1 - x - 3) = 0
<=> (5x - 1)(x - 2) = 0
<=> \(\orbr{\begin{cases}5x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0,2\\x=2\end{cases}}\)
Vậy x = 0,2 ; x = 2 là nghiệm phương trình
b) x3 - 5x2 - 3x + 15 = 0
<=> x2(x - 5) - 3(x - 5) = 0
<=> (x2 - 3)(x - 5) = 0
<=> \(\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x-5\right)=0\)
<=> \(x-\sqrt{3}=0\text{ hoặc }x+\sqrt{3}=0\text{ hoặc }x-5=0\)
<=> \(x=\sqrt{3}\text{hoặc }x=-\sqrt{3}\text{hoặc }x=5\)
Vậy \(x\in\left\{\sqrt{3};\sqrt{-3};5\right\}\)là giá trị cần tìm
c) (x - 3)2 - (5 - 2x)2 = 0
<=> (x - 3 + 5 - 2x)(x - 3 - 5 + 2x) = 0
<=> (-x + 2)(3x - 8) = 0
<=> \(\orbr{\begin{cases}-x+2=0\\3x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{3}\end{cases}}\)
Vậy tập nghiệm phương trình \(S=\left\{2;\frac{8}{3}\right\}\)
d) x3 + 4x2 + 4x = 0
<=> x(x2 + 4x + 4) = 0
<=> x(x + 2)2 = 0
<=> \(\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
Vậy tập nghiệm phương trình S = \(\left\{0;-2\right\}\)
a. 3.(x-2)+2.(x-3)=13
x=5
b. (x+1).(2-x)-(3x+5).(x+2)=-4x2+1
x=-9/10
c.x.(5-2x)+2x.(x-1)=13
x=13/3
d. (2x+3)2-(x-1)2=0
x=-2/3
e. x2.(3x-2)-8+12=0
x vô ngiệm
f x2+x=0
x=-1
g. x3-5x=0
x=0
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
a) \(3\left(x-2\right)+2\left(x-3\right)=1\)\(3\)
\(3x-6+2x-6=13\)
\(5x=13+6+6\)
\(5x=25\)
\(x=25\)
c) \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
d) \(\left(2x+3\right)^2-\left(x-1\right)^2=0\)
\(\left(2x+3-x+1\right)\left(2x+3+x-1\right)=0\)
\(\left(x+4\right)\left(3x+2\right)=0\)
\(\orbr{\begin{cases}x+4=0\\3x+2=0\end{cases}}=>\orbr{\begin{cases}x=-4\\x=\frac{-2}{3}\end{cases}}\)
f) \(x^2+x=0\)
\(x\left(x+1\right)=0\)
\(=>\orbr{\begin{cases}x=0\\x+1=0\end{cases}=>\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
g) \(x^3-5x=0\)
\(x^2\left(x-5\right)=0\)
\(=>\orbr{\begin{cases}x^2=0\\x-5=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=0\\x=5\end{cases}}\) \(\)
\(\)
Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
a) Ta có: \(2-x=2\left(x-2\right)^3\)
\(\Leftrightarrow-\left(x-2\right)-2\left(x-2\right)^3=0\)
\(\Leftrightarrow\left(x-2\right)\left[1+2\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
b) Ta có: \(8x^3-72x=0\)
\(\Leftrightarrow8x\left(x^2-9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy: S={0;3;-3}
c) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)
\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)
\(\Leftrightarrow x-1.5=0\)
hay x=1,5
d) Ta có: \(2x^3+3x^2+3+2x=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
hay \(x=-\dfrac{3}{2}\)
e) Ta có: \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
Vậy: S={0;1;-2}
f) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)
Vậy: S={0;2;12}