\(\sqrt{5}\cdot\sqrt{45}=\sqrt{5\cdot45}=\sqrt{225}=15\)

\(\sqrt{75\cdot48}=\sqrt{75}\cdot\sqrt{48}=5\sqrt{3}\cdot4\sqrt{3}=20\cdot3=60\)

\(\sqrt{2}\cdot\sqrt{162}=\sqrt{2\cdot162}=\sqrt{324}=18\)

\(\sqrt{0,25\cdot196}=\sqrt{0,25}\cdot\sqrt{196}=0,5\cdot14=7\)

\(\sqrt{17^2-8^2}=\sqrt{\left(17-8\right)\left(17+8\right)}=\sqrt{9\cdot25}=\sqrt{9}\cdot\sqrt{25}=3\cdot5=15\)

\(\sqrt{6,8^2-3,2^2}=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}=\sqrt{3,6\cdot10}=\sqrt{36}=6\)

\(\sqrt{36a^2}=\left|6a\right|=-6a\left(a< 0\right)\)

\(\sqrt{a^4b^6c^2}=\left|a^2b^3c\right|=a^2\left|b^3c\right|\)

\(\sqrt{4\left(a-3\right)^2}=2\left|a-3\right|=2a-6\left(a\ge3\right)\)

\(\sqrt{9\left(b-2\right)^2}=3\left|b-2\right|=6-3b\left(b< 2\right)\)

\(\sqrt{b^2\left(b-1\right)^2}=\left|b\left(b-1\right)\right|=b^2-b\left(b< 0\right)\)

ĐK : x ≥ -3/2

\(\Leftrightarrow x^2+4x+3-\left(2\sqrt{2x+3}-2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-\frac{8x+12-4}{2\left(\sqrt{2x+3}+1\right)}=0\)

\(\Leftrightarrow\left(x+1\right)\left[\left(x+3\right)-\frac{4}{\sqrt{2x+3}+1}\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+3\right)-\frac{4}{\sqrt{2x+3}+1}=0\end{cases}}\)

TH1 : x + 1 = 0 <=> x = -1 (tm) (1)

TH2 : \(\left(x+3\right)-\frac{4}{\sqrt{2x+3}+1}=0\Leftrightarrow\left(x+3\right)\sqrt{2x+3}+x+3=4\)

\(\Leftrightarrow\left(x+3\right)\left(\sqrt{2x+3}-1\right)+2x+2=0\)

\(\Leftrightarrow\left(x+3\right)\frac{2x+2}{\sqrt{2x+3}+1}+2x+2=0\)

\(\Leftrightarrow2\left(x+1\right)\left[\frac{x+3}{\sqrt{2x+3}+1}+1\right]=0\)(*)

Dễ thấy với x ≥ -3/2 thì \(\frac{x+3}{\sqrt{2x+3}+1}+1>0\)

nên (*) <=> x + 1 = 0 <=> x = -1 (tm) (2)

Từ (1) và (2) => pt có nghiệm x = -1

\(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{3\sqrt{3}}{\sqrt{2}}+\frac{2\sqrt{2}}{\sqrt{3}}-\frac{4\sqrt{3}}{\sqrt{2}}\)

\(=\frac{9+4-12}{\sqrt{6}}=\frac{1}{\sqrt{6}}\)

\(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)

\(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9\)

gọi 3 số đó là a,b,c

a+b+c=100

theo bdt cosi: a+b+c>=\(3\sqrt[3]{abc}\)

\(\Leftrightarrow100\ge3\sqrt[3]{abc}\Leftrightarrow\frac{1000000}{27}\ge abc\)

vậy abc đạt gtln là 1000000/27 hay tích 3 số đó có GTLN là 1000000/27

giải hệ pt    \(\hept{\begin{cases}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{cases}}\)

Giải

 \(\hept{\begin{cases}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{cases}}\)

\(\hept{\begin{cases}x^5+y^5=1\\x^4.x^5+y^4.y^5=x^4+y^4\end{cases}}\)

\(\hept{\begin{cases}x^5+y^5=1\\x^4.x^5+y^4.y^5-x^4-y^4=0\end{cases}}\)

\(\hept{\begin{cases}x^5=1-y^5\\x^4.\left(x^5-1\right)+y^4.\left(y^5-1\right)=0\end{cases}}\)

\(\hept{\begin{cases}x^5=1-y^5\\x^4.\left(1-y^5-1\right)+y^4.\left(y^5-1\right)=0\end{cases}}\)

\(\hept{\begin{cases}x^5=1-y^5\\x^4.\left(-y^5\right)+y^4.\left(-x^5\right)=0\end{cases}}\)

\(\hept{\begin{cases}x^5=1-y^5\\-x^4.y^5-y^4.x^5=0\end{cases}}\)

\(\hept{\begin{cases}x^5=1-y^5\\x^4.y^4\left(-y-x\right)=0\end{cases}}\)

...

\(yz\left(y+z\right)+xz\left(z-x\right)-xy\left(x+y\right)\)

\(=-[xy(x+y)-yz(y+z)-zx(z-x)]\)

\(=-(y.[x(x+y)-z(y+z)]-zx(z-x))\)

\(=-[y.(x^2+xy-zy-z^2)-zx(z-x)]\)

\(=-[y.(x^2-z^2+xy-zy)-zx(z-x)]\)

\(=-(y.[(x+z)(x-z)+y.(x-z)]-zx(z-x))\)

\(=-[y.(x-z)(x+z+y)+zx(x-z)]\)

\(=[(x-z)[y(x+z+y)+zx]]\)

\(=-(x-z)(yx+yz+y2+zx)\)

\(=-(x-z)(yx+zx+yz+y2)\)

\(=-[(x-z)[x.(y+z)+y.(y+z)]]\)

\(=-(x-z)(y+z)(x+y).\)

yz( y + z ) + xz ( z – x ) – xy ( x + y ) 

=(y+x)(z-x)(z+y)

nha bạn 

\(\hept{\begin{cases}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{cases}}\)

\(\Rightarrow x^9+y^9=\left(x^4+y^4\right)\left(x^5+y^5\right)=x^9+y^9+x^4y^5+x^5y^4\)

\(\Leftrightarrow x^4y^4\left(x+y\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0;y=0\\x+y=0\end{cases}}\)

Với \(x=0\Rightarrow y=1,y=0\Rightarrow x=1\).

Với \(x+y=0\Leftrightarrow x=-y\Leftrightarrow x^5=-y^5\Leftrightarrow x^5+y^5=0\)(mâu thuẫn với \(x^5+y^5=1\))

Vậy hệ có nghiệm là \(\left(0,1\right),\left(1,0\right)\).

sửa đề : \(\sqrt{2}x+2\sqrt{x}+4=0\)ĐK : x >= 0 

\(\Leftrightarrow x+\sqrt{x}+2=0\)

vì \(x+\sqrt{x}+2=x+\sqrt{x}+\frac{1}{4}+\frac{7}{4}=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{7}{4}>0\)

Vậy pt vô nghiệm 

đề của bạn ;-; \(\sqrt{2}x+2\sqrt{2}+4=0\Leftrightarrow x+2+2\sqrt{2}=0\Leftrightarrow x=-2-2\sqrt{2}\)

mình bổ sung đề của bạn ;-; \(x=-2-2\sqrt{2}\)(ktmx>=0)

Vậy pt vô nghiệm 

1, Để hs trên là hàm bậc nhất khi \(m\ne-1\)

Để hàm số trên đồng biến khi \(m+1>0\Leftrightarrow m>-1\)

Để hàm số trên khịch biến khi \(m+1< 0\Leftrightarrow m< -1\)

b, y = (m+1)x - 3 // y = 2x <=> \(\hept{\begin{cases}m+1=2\\-3\ne0\left(luondung\right)\end{cases}\Leftrightarrow m=1}\)

Thay m = 1 vào hs y = (m+1)x - 3 <=> y = 2x - 3 

bạn tự tìm tọa độ rồi tự vẽ nhé vd x = 1 => y = -1 ; x = 2 => y = 1 

Vậy A(1;-1) ; B(2;1) 

\(A=\frac{2}{\sqrt[3]{2}\left(\sqrt[3]{2}^2+\sqrt[3]{2}+1\right)}=\frac{2\left(\sqrt[3]{2}-1\right)}{\sqrt[3]{2}\left(\sqrt[3]{2}^2+\sqrt[3]{2}+1\right)\left(\sqrt[3]{2}-1\right)}=\frac{2\left(\sqrt[3]{2}-1\right)}{\sqrt[3]{2}}=2-\sqrt[3]{4}\)