(x – 5) (3 + 5x)(6x + 1) = 0
d) (x – 3)(3x + 5) = (2x - 6)(x + 1)
(2x2 + 1)(x2 - 2) = 0
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8 tháng 1 2021
a) Ta có: \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2+4\left(x^2-5x\right)+6\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x+4\right)+6\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-2x-3x+6\right)\left(x^2-x-4x+4\right)=0\)
\(\Leftrightarrow\left[x\left(x-2\right)-3\left(x-2\right)\right]\left[x\left(x-1\right)-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\)
Vậy: S={1;2;3;4}
b) Ta có: \(\left(2x+1\right)^2-2x-1=2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)
\(\Leftrightarrow\left(2x+1+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{2}\right\}\)
c) Ta có: \(x\left(x-1\right)\left(x^2-x+1\right)-6=0\)
\(\Leftrightarrow x\left(x^3-x^2+x-x^2+x-1\right)-6=0\)
\(\Leftrightarrow x\left(x^3-2x^2+2x-1\right)-6=0\)
\(\Leftrightarrow x^4-2x^3+2x^2-x-6=0\)
\(\Leftrightarrow x^4-2x^3+2x^2-4x+3x-6=0\)
\(\Leftrightarrow x^3\left(x-2\right)+2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+2x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x^2-1\right)+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x+3\right)=0\)
mà \(x^2-x+3>0\forall x\)
nên (x-2)(x+1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy: S={2;-1}
d) Ta có: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)^2+2x\left(x^2+1\right)+x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2+1+2x\right)+x\left(x^2+1+2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
25 tháng 3 2020
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
23 tháng 2 2022
Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
20 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`1.`
\(\left(-4xy\right)\cdot\left(2xy^2-3x^2y\right)\)
`=`\(\left(-4xy\right)\left(2xy^2\right)+\left(-4xy\right)\left(-3x^2y\right)\)
`=`\(-8\left(x\cdot x\right)\left(y\cdot y^2\right)+12\left(x\cdot x^2\right)\left(y\cdot y\right)\)
`=`\(-8x^2y^3+12x^3y^2\)
`2.`
\(\left(-5x\right)\left(3x^3+7x^2-x\right)\)
`=`\(\left(-5x\right)\left(3x^3\right)+\left(-5x\right)\left(7x^2\right)+\left(-5x\right)\left(-x\right)\)
`=`\(-15x^4-35x^3+5x^2\)
`3.`
\(\left(3x-2\right)\left(4x+5\right)-6x\left(2x-1\right)\)
`=`\(3x\left(4x+5\right)-2\left(4x+5\right)-12x^2+6x\)
`=`\(12x^2+15x-8x-10-12x^2+6x\)
`=`\(\left(12x^2-12x^2\right)+\left(15x-8x+6x\right)-10\)
`=`\(13x-10\)
`4.`
\(2x^2\left(x^2-7x+9\right)\)
`=`\(2x^2\cdot x^2+2x^2\cdot\left(-7x\right)+2x^2\cdot9\)
`=`\(2x^4-14x^3+18x^2\)
`5.`
\(\left(3x-5\right)\left(x^2-5x+7\right)\)
`=`\(3x\left(x^2-5x+7\right)-5\left(x^2-5x+7\right)\)
`=`\(3x^3-15x^2+21x-5x^2+25x-35\)
`=`\(3x^3-20x^2+46x-35\)
9 tháng 9 2021
a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)
\(\Leftrightarrow20x-5+2-6x-6x-30=10\)
\(\Leftrightarrow8x=43\)
hay \(x=\dfrac{43}{8}\)
b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)
\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)
\(\Leftrightarrow6x^2-3x-3=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
RH
18 tháng 2 2022
\(a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}\)
18 tháng 2 2022
d: \(\Leftrightarrow3x^2-6x-2x+4=0\)
=>(x-2)(3x-2)=0
=>x=2 hoặc x=2/3
e: \(\Leftrightarrow3x\left(x^2-2x-3\right)=0\)
=>x(x-3)(x+1)=0
hay \(x\in\left\{0;3;-1\right\}\)
f: \(\Leftrightarrow x^2-5x-2+x=0\)
\(\Leftrightarrow x^2-4x-2=0\)
\(\Leftrightarrow\left(x-2\right)^2=6\)
hay \(x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}\)
17 tháng 9 2021
\(1,\\ a,=3x^3-2x^2+5x\\ b,=2x^3y^2+\dfrac{2}{9}x^4y^2-\dfrac{1}{3}x^2y^3\\ c,=x^2-2x+6x-12=x^2+4x-12\\ 2,\\ a,\Rightarrow6x-9+4-2x=-3\\ \Rightarrow4x=2\Rightarrow x=\dfrac{1}{2}\\ b,\Rightarrow5x-2x^2+2x^2-2x=13\\ \Rightarrow3x=13\Rightarrow x=\dfrac{13}{3}\\ c,\Rightarrow5x^2-5x-5x^2+7x-10x+14=6\\ \Rightarrow-8x=-8\Rightarrow x=1\\ d,\Rightarrow6x^2+9x-6x^2+4x-15x+10=8\\ \Rightarrow-2x=-2\Rightarrow x=1\)
17 tháng 9 2021
\(3,\\ A=2x^2+x-x^3-2x^2+x^3-x+3=3\\ B=6x^2-10x+33x-55-6x^2-14x-9x-21=-76\)
P/S : Tại vì đề bài không cho biết tìm x hay giải phương trình nên mình sẽ giải phương trình luôn nhé
(x - 5)(3 + 5x)(6x + 1) = 0 <=> \(\hept{\begin{cases}x-5=0\\3+5x=0\\6x+1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=5\\x=-\frac{3}{5}\\x=-\frac{1}{6}\end{cases}}\)
Vậy \(S=\left\{5;-\frac{3}{5};-\frac{1}{6}\right\}\)
(x - 3)(3x + 5) = (2x - 6)(x + 1)
<=> 3x2 + 5x - 9x - 15 = 2x2 + 2x - 6x - 6
<=> 3x2 + 5x - 9x - 15 - 2x2 - 2x + 6x + 6 = 0
<=> x2 - 9 = 0
<=> (x - 3)(x + 3) = 0
<=> \(\hept{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy \(S=\left\{3;-3\right\}\)
(2x2 + 1)(x2 - 2) = 0
Vì 2x2 + 1 > 0 nên không tìm được x
x2 - 2 = 0 <=> \(x^2-\left(\sqrt{2}\right)^2=0\)
<=> \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
<=> \(\hept{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)
Vậy \(S=\left\{\sqrt{2};-\sqrt{2}\right\}\)
a, (x – 5) (3 + 5x) (6x + 1) = 0
\(\Rightarrow\)Hoặc x - 5 = 0 hoặc 3 + 5x = 0 hoặc 6x + 1 = 0
\(\Leftrightarrow\)Hoặc x = 5 hoặc x = -3/5 hoặc x = -1/6
b, (x – 3)(3x + 5) = (2x - 6)(x + 1)
\(\Leftrightarrow\)(x – 3) (3x + 5) - (2x - 6) (x + 1) = 0
\(\Leftrightarrow\)\(3x^2+5x-9x-15-2x^2-2x+6x+6=0\)
\(\Leftrightarrow x^2-9=0\)
\(\Leftrightarrow\)( x - 3 ) ( x + 3) = 0
\(\Leftrightarrow\)Hoặc x = 3 hoặc x = -3
c, (2x2 + 1)(x2 - 2) = 0
\(\Leftrightarrow\)Hoặc \(2x^2+1=0\)hoặc \(x^2-2=0\)
\(\Leftrightarrow\)Hoặc \(2x^2=-1\)(vô lí) hoặc \(x^2=2\)
\(\Leftrightarrow\)\(x=\sqrt{2}\) hoặc \(x=-\sqrt{2}\)
\(\Leftrightarrow\)