A=5x(2-X )2 +3x |2-X |=0
B= -|3X+7| +9
* x là nhân
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a)(2x-3)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)
Vậy x=3/2 hoặc x=-5
a) \(\left(2x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};-5\right\}\)
b) \(3x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là: \(S=\left\{2;\dfrac{7}{2}\right\}\)
c) \(5x\left(2x-3\right)-6x+9=0\)
\(\Leftrightarrow5x\left(2x-3\right)-3\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\5x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};\dfrac{3}{5}\right\}\)
a: \(\left(x+5\right)^2-\left(x-5\right)^2-2x+1=0\)
=>\(x^2+10x+25-\left(x^2-10x+25\right)-2x+1=0\)
=>\(x^2+8x+26-x^2+10x-25=0\)
=>18x+1=0
=>\(x=-\dfrac{1}{18}\)
b: \(\left(2x-7\right)^2-\left(x+3\right)^2=3x^2+6\)
=>\(4x^2-28x+49-\left(x^2+6x+9\right)-3x^2-6=0\)
=>\(x^2-28x+43-x^2-6x-9=0\)
=>34-34x=0
=>34x=34
=>x=1
c: \(\left(3x+2\right)^2-9\left(x-5\right)\left(x+5\right)=225-5x\)
=>\(9x^2+12x+4-9\left(x^2-25\right)-225+5x=0\)
=>\(9x^2+17x+4-225-9x^2+225=0\)
=>17x+4=0
=>x=-4/17
a. 2x\(^2\)-8=0
2x\(^2\)=8
x\(^2\)=4
x=2
b.3x\(^3\)-5x=0
x(3x\(^2\)-5)=0
\(\left[{}\begin{matrix}x=0\\x^2-5=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=0\\x^2=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=^+_-\sqrt{5}\end{matrix}\right.\)
c.x\(^4\)+3x\(^2\)-4=0\(^{\left(\cdot\right)}\)
đặt t=x\(^2\) (t>0)
ta có pt: t\(^2\)+3t-4=0 \(^{\left(1\right)}\)
thấy có a+b+c=1+3+(-4)=0 nên pt\(^{\left(1\right)}\) có 2 nghiệm
t\(_1\)=1; t\(_2\)=\(\dfrac{c}{a}\)=-4
khi t\(_1\)=1 thì x\(^2\)=1 ⇒x=\(^+_-\)1
khi t\(_2\)=-4 thì x\(^2\)=-4 ⇒ x=\(^+_-\)2
vậy pt đã cho có 4 nghiệm x=\(^+_-\)1; x=\(^+_-\)2
d)3x\(^2\)+6x-9=0
thấy có a+b+c= 3+6+(-9)=0 nên pt có 2 nghiệm
x\(_1\)=1; x\(_2\)=\(\dfrac{c}{a}=\dfrac{-9}{3}=-3\)
e. \(\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}\) (ĐK: x#5; x#2 )
⇔\(\dfrac{\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}+\dfrac{3\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}\)=\(\dfrac{6\left(x-5\right)}{\left(x-5\right)\left(2-x\right)}\)
⇒2x - x\(^2\) + 4 - 2x + 6x - 6x\(^2\) + 12 - 6x - 6x +30 = 0
⇔-7x\(^2\) - 6x + 46=0
Δ'=b'\(^2\)-ac = (-3)\(^2\) - (-7)\(\times\)46= 9+53 = 62>0
\(\sqrt{\Delta'}=\sqrt{62}\)
vậy pt có 2 nghiệm phân biệt
x\(_1\)=\(\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{3+\sqrt{62}}{-7}\)
x\(_2\)=\(\dfrac{-b'-\sqrt{\Delta'}}{a}=\dfrac{3-\sqrt{62}}{-7}\)
vậy pt đã cho có 2 nghiệm x\(_1\)=.....;x\(_2\)=......
câu g làm tương tự câu c
a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
c: Ta có: \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b.
PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$
$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$
$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$
$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$
$\Leftrightarrow (x-3)^2(2x+1)^2=0$
$\Leftrightarrow (x-3)(2x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $2x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$
d.
$x^2-2x=24$
$\Leftrightarrow x^2-2x-24=0$
$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$
$\Leftrightarrow x=-4$ hoặc $x=6$
a: =>3x^2-3x-2x+2=0
=>(x-1)(3x-2)=0
=>x=2/3 hoặc x=1
b: =>2x^2=11
=>x^2=11/2
=>\(x=\pm\dfrac{\sqrt{22}}{2}\)
c: Δ=5^2-4*1*7=25-28=-3<0
=>PTVN
f: =>6x^4-6x^2-x^2+1=0
=>(x^2-1)(6x^2-1)=0
=>x^2=1 hoặc x^2=1/6
=>\(\left[{}\begin{matrix}x=\pm1\\x=\pm\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)
d: =>(5-2x)(5+2x)=0
=>x=5/2 hoặc x=-5/2
e: =>4x^2+4x+1=x^2-x+9 và x>=-1/2
=>3x^2+5x-8=0 và x>=-1/2
=>3x^2+8x-3x-8=0 và x>=-1/2
=>(3x+8)(x-1)=0 và x>=-1/2
=>x=1
Bài 2
a) 5x² + 30y
= 5(x² + 6y)
b) x³ - 2x² - 4xy² + x
= x(x² - 2x - 4y² + 1)
= x[(x² - 2x + 1) - 4y²]
= x[(x - 1)² - (2y)²]
= x(x - 1 - 2y)(x - 1 + 2y)
Bài 3:
a: \(2x\left(x-3\right)-x+3=0\)
=>\(2x\left(x-3\right)-\left(x-3\right)=0\)
=>(x-3)(2x-1)=0
=>\(\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
b: \(\left(3x-1\right)\left(2x+1\right)-\left(x+1\right)^2=5x^2\)
=>\(6x^2+3x-2x-1-x^2-2x-1=5x^2\)
=>\(5x^2-x-2=5x^2\)
=>-x-2=0
=>-x=2
=>x=-2
a: Ta có: \(3x\left(3x-1\right)-\left(3x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow9x^2-3x-9x^2+1=0\)
\(\Leftrightarrow3x=1\)
hay \(x=\dfrac{1}{3}\)
b: Ta có: \(x^2-5x+25-5x=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\)
hay x=5
a, 3x - 7 = 0
<=> 3x = 7
<=> x = 7/3
b, 8 - 5x = 0
<=> -5x = -8
<=> x = 8/5
c, 3x - 2 = 5x + 8
<=> -2x = 10
<=> x = -5
e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)
Câu hỏi là j vậy bn ?