\(A=x^2+4x^4\)

\(\Rightarrow A=\left(2x^2\right)^2+4x^3+\left(x\right)^2-4x^3\)

\(\Rightarrow\left(2x^2+x\right)^2-4x^3\)

=> Ko là số chính phương

\(B=y^2-12y+36\)

\(B=y^2-2.6y+6^2\)

\(\Rightarrow B=\left(y-6\right)^2\)

=> Là số chính phương

a/ \(\left\{a\right\};\left\{b\right\};\left\{a;b\right\};\varnothing\)

b/ \(\left\{1\right\};\left\{2\right\};\left\{3\right\};\left\{1;2\right\};\left\{1;3\right\};\left\{2;3\right\};\left\{1;2;3\right\};\varnothing\)

c/ \(\left\{0\right\};\left\{1\right\};\left\{2\right\};\left\{3\right\};\left\{0;1\right\};\left\{0;2\right\};\left\{0;3\right\};\left\{1;2\right\};\left\{1;3\right\};\left\{2;3\right\};\left\{0;1;2\right\};\left\{1;2;3\right\};\left\{0;2;3\right\};\left\{0;1;3\right\};\left\{0;1;2;3\right\};\varnothing\)

d/ \(\left\{1\right\};\left\{-2\right\};\left\{1;-2\right\};\varnothing\)

1. Tìm x, biết :

a. ( x - \(\frac{3}{4}\)) \(^2\)= 0

=> x - \(\frac{3}{4}\)= 0

=> x = 0 + \(\frac{3}{4}\)

=> x = \(\frac{3}{4}\)

b. ( x + \(\frac{1}{2}\)) \(^2\)= \(\frac{9}{64}\)

=> ( x + \(\frac{1}{2}\)) \(^2\)= ( \(\frac{3}{8}\)) \(^2\)

=> x + \(\frac{1}{2}\)= \(\frac{3}{8}\)

=> x = \(\frac{3}{8}\)- \(\frac{1}{2}\)

=> x = \(\frac{-1}{8}\)

c.  \(\frac{\left(-2\right)^x}{16}=-8\)

=> \(\frac{\left(-2\right)^x}{16}=\frac{-8}{1}=\frac{-128}{16}\)

=> ( -2)\(^x\)= -128

=> ( -2 ) \(^x\)= ( -2) \(^7\)

=> x = 7

\(A=\left\{x\in R|\left(x-2x^2\right)\left(x^2-3x+2\right)=0\right\}\)

Giải phương trình sau :

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

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

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

\(\Rightarrow A=\left\{0;\dfrac{1}{2};1;2\right\}\)

\(B=\left\{n\in N|3< n\left(n+1\right)< 31\right\}\)

Giải bất phương trình sau :

\(3< n\left(n+1\right)< 31\)

\(\Leftrightarrow\left\{{}\begin{matrix}n\left(n+1\right)>3\\n\left(n+1\right)< 31\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n^2+n-3>0\\n^2+n-31< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n< \dfrac{-1-\sqrt[]{13}}{2}\cup n>\dfrac{-1+\sqrt[]{13}}{2}\\\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1-\sqrt[]{13}}{2}\\\dfrac{-1+\sqrt[]{13}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)

Vậy \(B=\left(\dfrac{-1-5\sqrt[]{5}}{2};\dfrac{-1-\sqrt[]{13}}{2}\right)\cup\left(\dfrac{-1+\sqrt[]{13}}{2};\dfrac{-1+5\sqrt[]{5}}{2}\right)\)

\(\Rightarrow A\cap B=\left\{2\right\}\)

Vì trong biểu thức phải có 1 số =0 thì kết quả mới bằng 0

Nên bạn xét từng trường hơp nó bằng 0 nha

mk ko bt trình bày đầy đủ nên chỉ bt vậy thui

tik mk nha

\(\left(x+2\right)\left(x+4\right)\left(x^2-9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+4=0\\x^2-9=0\Rightarrow x^2=9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\\\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\end{matrix}\right.\)

Vậy................

Vì \(x\in N\) nên x>=0
\(\Rightarrow x+2>=2>0\)
\(x+4>=4>0\)
\(x+3>=3>0\)
Do đó \(\left(x+2\right)\left(x+4\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)

KL: x = 3

Câu 1:

\(2xyz=1-\left(x+y+z\right)+xy+yz+zx\)

\(\Rightarrow xy+yz+zx=2xyz+\left(x+y+z\right)-1\)

\(VT=x^2+y^2+z^2=\left(x+y+z\right)^2-2\left(xy+yz+zx\right)\)

\(=\left(x+y+z\right)^2-2\left(x+y+z\right)-4xyz+2\)

\(VT\ge\left(x+y+z\right)^2-2\left(x+y+z\right)-\frac{4}{27}\left(x+y+z\right)^3+2\)

\(VT\ge\frac{4}{27}\left[\frac{15}{4}-\left(x+y+z\right)\right]\left(x+y+z-\frac{3}{2}\right)^2+\frac{3}{2}\ge\frac{3}{2}\)

(Do \(0< x;y;z< 1\Rightarrow x+y+z< 3< \frac{15}{4}\))

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)

Câu 2:

Từ điều kiện bài này có thể đặt ẩn phụ và AM-GM ra luôn kết quả, nhưng hơi rắc rối khi người ta hỏi từ đâu mà có cách đặt ẩn phụ như vậy, do đó ta giải trâu :D

\(x^2+y^2+z^2+xyz=4\)

\(\Leftrightarrow\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{4}+2\left(\frac{x}{2}.\frac{y}{z}.\frac{z}{2}\right)=1\)

\(\Leftrightarrow\frac{xy}{2z}.\frac{xz}{2y}+\frac{xy}{2z}.\frac{yz}{2x}+\frac{yz}{2x}.\frac{xz}{2y}+2\left(\frac{xy}{2z}.\frac{yz}{2x}.\frac{xy}{2y}\right)=1\)

Đặt \(\left(\frac{xy}{2z};\frac{zx}{2y};\frac{yz}{2x}\right)=\left(m;n;p\right)\Rightarrow mn+np+pn+2mnp=1\)

\(\Leftrightarrow2\left(n+1\right)\left(m+1\right)\left(p+1\right)=\left(n+1\right)\left(m+1\right)+\left(n+1\right)\left(p+1\right)+\left(m+1\right)\left(p+1\right)\)

\(\Leftrightarrow\frac{1}{n+1}+\frac{1}{m+1}+\frac{1}{p+1}=2\)

\(\Leftrightarrow1=\frac{n}{n+1}+\frac{m}{m+1}+\frac{p}{p+1}\ge\frac{\left(\sqrt{n}+\sqrt{m}+\sqrt{p}\right)^2}{m+n+p+3}\)

\(\Leftrightarrow m+m+p+2\left(\sqrt{mn}+\sqrt{np}+\sqrt{mp}\right)\le m+n+p+3\)

\(\Leftrightarrow\sqrt{mn}+\sqrt{np}+\sqrt{mp}\le\frac{3}{2}\)

\(\Leftrightarrow\frac{x}{2}+\frac{y}{2}+\frac{z}{2}\le\frac{3}{2}\Leftrightarrow x+y+z\le3\)

@Nguyễn Việt Lâm, @Akai Haruma, @tth_new

giúp em vs ạ! e cảm ơn nhiều!

a: =>5x+1=6/7 hoặc 5x+1=-6/7

=>5x=-1/7 hoặc 5x=-13/7

=>x=-1/35 hoặc x=-13/35

b: =>x-2/9=4/9

=>x=6/9=2/3

c: =>8x+1=5

=>8x=4

hay x=1/2