Do \(\left|x\right|,\left|x^2+x\right|\ge0\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x^2+x=0\end{matrix}\right.\)

\(\Rightarrow x=0\)

x = 0

Vì bất cứ số nào nhân hay chia với o đều = 0

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)+5x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(6x+3\right)=0\\ \Leftrightarrow3\left(x+2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

bạn ơi cho mình hỏi là 6x + 3 sao ra v đc bạn

\(\left(3x-1\right)^2.\left(x+5\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)

Bạn làm đầy đủ hơn được không ạ? 

cj cs thể ghi lại phần kết luận đc ko ạ em ko nhìn rõ

bn ko nói là tìm x hay y hay xy à

`xy-x-y=0`

`<=>xy-x-y+1=1`

`<=> x(y-1)-(y-1)=1`

`<=> (y-1)(x-1)=1`

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y-1=1\\x-1=1\end{matrix}\right.\\\left\{{}\begin{matrix}y-1=-1\\x-1=-1\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2\\x=2\end{matrix}\right.\\\left\{{}\begin{matrix}y=0\\x=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow2\left(x^2-\dfrac{3}{2}x+\dfrac{5}{2}\right)=0\\ \Leftrightarrow2\left(x-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}+\dfrac{31}{16}\right)=0\\ \Leftrightarrow2\left(x-\dfrac{3}{4}\right)^2+\dfrac{31}{8}=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-\dfrac{3}{4}\right)^2+\dfrac{31}{8}\ge\dfrac{31}{8}>0\right]\)

cảm ơn nhé

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

\(\left(x-3\right)\left(x^2+5\right)=0\)

\(\Rightarrow x-3=0\) hoặc \(x^2+5=0\)

      x= 3                   \(x^2=-5\) ( vô lý- do \(x^2\ge0\) với mọi x)

Vậy ...

a: \(\left(5\sin x-2\right)\cdot cos^2x=3\cdot\sin^2x-3\cdot\sin^3x\)

=>\(\left(5\cdot\sin x-2\right)\left(1-\sin^2x\right)=3\cdot\sin^2x-3\cdot\sin^3x\)

=>\(5\cdot\sin x-5\cdot\sin^3x-2+2\cdot\sin^2x-3\cdot\sin^2x+3\cdot\sin^3x=0\)

=>\(-2\cdot\sin^3x-\sin^2x+5\cdot\sin x-2=0\)

=>\(2\cdot\sin^3x+\sin^2x-5\cdot\sin x+2=0\)

=>\(2\cdot\sin^3x-2\cdot\sin^2x+3\cdot\sin^2x-3\cdot\sin x-2\cdot\sin x+2=0\)

=>(sin x-1)(\(2\cdot\sin^2x+3\cdot\sin x-2\) )=0

=>\(\left(\sin x-1\right)\left(\sin x+2\right)\left(2\cdot\sin x-1\right)=0\)

=>(sin x-1)(2sin x-1)=0

TH1: sin x-1=0

=>sin x=1

=>\(x=\frac{\pi}{2}+k2\pi\)

TH2: 2 sin x-1=0

=>sin x=1/2

=>\(\left[\begin{array}{l}x=\frac{\pi}{6}+k2\pi\\ x=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\)

b: \(\sin7x\cdot\sin10x+\sin6x\cdot\sin3x-\sin4x\cdot\sin x=0\)

=>\(\frac12\cdot\left\lbrack cos\left(10x-7x\right)-cos\left(10x+7x\right)\right\rbrack+\frac12\left\lbrack cos\left(6x-3x\right)-cos\left(6x+3x\right)\right\rbrack-\frac12\cdot\left\lbrack cos\left(4x-x\right)-cos\left(4x+x\right)\right\rbrack=0\)

=>cos3x-cos17x+cos3x-cos9x-cos3x+cos5x=0

=>-cos17x-cos9x+cos3x+cos5x=0

=>\(cos17x-cos5x+cos9x-cos3x=0\)

=>\(-2\cdot\sin\frac{17x+5x}{2}\cdot\sin\frac{17x-5x}{2}-2\cdot\sin\frac{9x+3x}{2}\cdot\sin\frac{9x-3x}{2}=0\)

=>\(\sin11x\cdot\sin6x+\sin6x\cdot\sin3x=0\)

=>sin6x(sin11x+sin3x)=0

TH1: sin 6x=0

=>\(6x=k\pi\)

=>\(x=\frac{k\pi}{6}\)

TH2: sin 11x+sin 3x=0

=>sin 11x=-sin 3x=sin(-3x)

=>\(\left[\begin{array}{l}11x=-3x+k2\pi\\ 11x=\pi+3x+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}14x=k2\pi\\ 8x=\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{k\pi}{7}\\ x=\frac{\pi}{8}+\frac{k\pi}{4}\end{array}\right.\)

Ta có: \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\)

\(\left(y+0.4\right)^{100}\ge0\forall y\)

\(\left(z-3\right)^{678}\ge0\forall z\)

Do đó: \(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0.4\right)^{100}+\left(z-3\right)^{678}\ge0\forall x,y,z\)

Dấu '=' xảy ra khi

\(\left\{{}\begin{matrix}x-\dfrac{1}{5}=0\\y+0.4=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{2}{5}\\z=3\end{matrix}\right.\)

Vậy: (x,y,z)=\(\left(\dfrac{1}{5};-\dfrac{2}{5};3\right)\)

\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...