Bài 2: 

a: =>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x.y}{2.3}=\dfrac{54}{6}=9\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=81\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm9\end{matrix}\right.\)

b) \(\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\left(\dfrac{x}{5}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x^2-y^2}{5^2-3^2}=\dfrac{4}{16}=\dfrac{1}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{25}{4}\\y^2=\dfrac{9}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{5}{2}\\y=\pm\dfrac{3}{2}\end{matrix}\right.\)

c: Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\)

nên \(\dfrac{x}{10}=\dfrac{y}{15}\)

Ta có: \(\dfrac{y}{5}=\dfrac{z}{7}\)

nên \(\dfrac{y}{15}=\dfrac{z}{21}\)

mà \(\dfrac{x}{10}=\dfrac{y}{15}\)

nên \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{92}{46}=2\)

Do đó: x=20; y=30; z=42

a) \(x\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b) \(\left(-7-x\right)\left(-x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)

c) \(\left(x+3\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

d) \(\left(x-3\right)\left(x^2+12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)

\(\Rightarrow x=3\)

e) \(\left(x+1\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow-1\le x\le2\)

f) \(\left(x-3\right)\left(x-5\right)\le0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow3\le x\le5\)

a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3

c) => \(\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

$A\geq \frac{9}{x+2+y+2+z+2}=\frac{9}{x+y+z+6}$

Áp dụng BĐT Bunhiacopxky:

$(x^2+y^2+z^2)(1+1+1)\geq (x+y+z)^2$

$\Rightarrow 9\geq (x+y+z)^2\Rightarrow x+y+z\leq 3$

$\Rightarrow A\geq \frac{9}{x+y+z+6}\geq \frac{9}{3+6}=1$
Vậy $A_{\min}=1$. Dấu "=" xảy ra khi $x=y=z=1$

a) |x + 25| + |-y + 5| =0

=> |x + 25| = 0 hoặc |-y + 5| = 0

Từ đó bạn cứ bỏ giá trị tuyệt đối rồi tính nha! Mấy bài khác cũng vậy

a) (x-1)(x-2)=0

x-1=0 --> x = 1

x-2=0 --> x = 2

d) x^2(x + 1) + 27(x + 1)=0

(x+1)(x^2+27)=0

x+1=0 --> x = -1

x^2+27=0 (vô lí)

a) \(x\left(x-2\right)-x+2=0\)

\(x\left(x-2\right)-\left(x-2\right)=0\)

\(\left(x-1\right)\left(x-2\right)=0\)

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

a) 

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

b) \(\left(3-2x\right)^2-\left(x-1\right)^2=0\)

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

\(\left(3-3x+1\right)\left(3-x-1\right)=0\)

TH1:3-3x+1=0⇒x\(=\dfrac{4}{3}\)

TH2:3-x-1=0⇒x=2

Đề bài yêu cầu gì vậy em.