Toán

\(P=\left|\sqrt{x-3}+3\right|+\left|\sqrt{x-3}+1\right|\)

\(=2\sqrt{x-3}+4>=4\)

Dấu = xảy ra khi x=3

`A=(x+4)^2-(4-x)^2`

`A=(x+4-4+x)(x+4+4-x)`

`A=2x.8=16x`

Thay `x=15` vào `A` có: `A=16.15=240`

`A=(x+4)^2-(4-x)^2`

`A=(x^2 +8x+16)-(x^2-8x+16)`

`A=x^2+8x+16-x^2+8x-16`

`A=16x `

Thay `x=15`

`A=16.15`

`A=240`

\(A=\left(x+4\right)^2-\left(4-x\right)^2\) tại \(x=15\) 

Thay: \(x=15\) vào \(A=\left(x+4\right)^2-\left(4-x\right)^2\) 

Ta có: \(A=\left(15+4\right)^2-\left(4-15\right)^2\\ =\left(19\right)^2-\left(-11\right)^2\\ =361-121\\ =240.\)

`a)A=0`

`=>x(x-1/2)=0`

`@TH1:x=0`

`@TH2:x-1/2=0=>x=1/2`

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`b)A < 0`

`=>x(x-1/2) < 0`

`@TH1:{(x < 0),(x-1/2 > 0):}=>{(x < 0),(x > 1/2):}=>x` không có giá trị t/m

`@TH2:{(x > 0),(x-1/2 < 0):}=>{(x > 0),(x < 1/2):}=>0 < x < 1/2`

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`c)A > 0`

`=>x(x-1/2) > 0`

`@TH1:{(x > 0),(x-1/2 > 0):}=>{(x > 0),(x > 1/2):}=>x > 1/2`

`@TH2:{(x < 0),(x-1/2 < 0):}=>{(x < 0),(x < 1/2):}=>x < 0`

`a, A = 0 => x(x-1/2) = 0`

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

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

`b, A < 0`

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

`->` \(\left[{}\begin{matrix}\left[{}\begin{matrix}x>0\\x< \dfrac{1}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x>\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

`=>` x < 1/2`

`c, A > 0`

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

`=>` \(\left[{}\begin{matrix}\left[{}\begin{matrix}x>0\\x>\dfrac{1}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x< \dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

`=> x > 1/2` hoặc `x < 0`

Trong tứ giác ABCD, có:

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\left(...\right)\)

\(\Rightarrow2\widehat{B}+\widehat{B}+3\widehat{D}+\widehat{D}=360^0\)

\(\Rightarrow3\widehat{B}+4\widehat{D}=360^0\)

Có: \(360:3=120\), \(360:4=90\)

\(\Rightarrow3.120^0+4.90^0=360^0.2\)

\(\Rightarrow3.60^0+4.45^0=360^0\)

\(\Rightarrow\widehat{A}+2\widehat{B}=180^0;\widehat{C}+3\widehat{D}=180^0\)

\(\Rightarrow\widehat{A}=120^0,\widehat{B}=600^0,\widehat{C}=135^0,\widehat{D}=45^0\)

a) \(\left(\dfrac{3}{5}\right)^{15}:\left(\dfrac{27}{125}\right)^5=\left(\dfrac{3}{5}\right)^{15}:\left[\left(\dfrac{3}{5}\right)^3\right]^5=\left(\dfrac{3}{5}\right)^{15}:\left(\dfrac{3}{5}\right)^{15}=1\)

b) \(\left(\dfrac{1}{7}\right)^4.\dfrac{1}{7}.49^3=\left(\dfrac{1}{7}\right)^5=\left(7^2\right)^3=\left(\dfrac{1}{7}\right)^5.7^6=7\)

\(a;b;c>0\) \((a;b;c\ge0\) \(là\) \(sai)\)

\(\Sigma\dfrac{4}{\left(a+b\right)^3}\ge\Sigma\dfrac{c}{a+b}\Leftrightarrow\Sigma\dfrac{4}{\left(3-c\right)^3}\ge\Sigma\dfrac{c}{3-c}\left(1\right)\)

\(\dfrac{4}{\left(3-c\right)^3}\ge\dfrac{c}{3-c}\Leftrightarrow4\left(3-c\right)\ge c\left(3-c\right)^3\Leftrightarrow4\left(c-3\right)-c\left(3-c\right)^3\ge0\Leftrightarrow-\left(3-c\right)\left(c-4\right)\left(c-1\right)^2\ge0\left(2\right)\)

\(do:a,b,c>0;a+b+c=3\Rightarrow0< a,b,c< 3\Rightarrow\left(2\right)\) \(đúng\)

\(tương\) \(tự\Rightarrow\dfrac{4}{\left(3-a\right)^3}\ge\dfrac{a}{3-a};\dfrac{4}{\left(3-b\right)^3}\ge\dfrac{b}{3-b}\)

\(\Rightarrow\)\(\left(1\right)đúng\Rightarrowđpcm\)

a: X={2;3;6;7}

=>Có 1 tập

b: X={2;3;6;1;7;8;5}

X={1;7;8;5}

X={2;1;7;8;5}

X={3;1;7;8;5}

X={6;1;7;8;5}

X={2;3;1;7;8;5}

X={3;6;1;7;8;5}

X={3;6;1;7;8;5}

=>Có 8 tập

`sin (3x+20^o)=-1/2`

`<=>` $\left[\begin{matrix} 3x+20^o =-30^o +k360^o\\ 3x+20^o =210^o +k360^o\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=\dfrac{-50^o }{3}+k120^o \\ x=\dfrac{190^o }{3}+k120^o\end{matrix}\right.$   `(k in ZZ)`

Ta có :

\(sin\left(3x+\dfrac{\pi}{9}\right)=\dfrac{-\pi}{6}\)

⇔ \(\left[{}\begin{matrix}3x+\dfrac{\pi}{9}=\dfrac{-\pi}{6}+k2\pi\\3x+\dfrac{\pi}{9}=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}3x=\dfrac{-5\pi}{18}+k2\pi\\3x=\dfrac{19\pi}{18}+k2\pi\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=-\dfrac{5\pi}{54}+\dfrac{k2\pi}{3}\\x=\dfrac{19\pi}{54}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

Vậy ...

Phân số chỉ số phần bể đã có nước :

`3/5 + 2/9= 37/45(phần-bể)`

Phân số chỉ số phần bể đã được vòi thứ `3` chảy :

`1-37/45 = 8/45(phần-bể)`

Bể đó chứa được số lít nước :

`64 :8/45=  360(lít)`

D/s....

`= 7^5 + 7^9 . (5^4 + 5^6) . 0`

`= 7^5`

=7⁵+7⁹.(5⁴+5⁶).0

=7⁵+0

=7⁵