a = -3 nha bạn!

888653222,22222,4446,1234,-6,-333,-407639674.

a) M = -x² - 4x + 2 = -x² - 4x - 4 + 6 = -( x² + 4x + 4 ) + 6 = -( x + 2 )² + 6

\(-\left(x+2\right)^2\le0\forall x\Rightarrow-\left(x+2\right)^2+6\le6\)

Dấu " = " xảy ra <=> x + 2 = 0 => x = -2

Vậy M_Max = 6 , đạt được khi x = -2

b) N = -2y² - 3y + 5 = -2( y² + 3/2y + 9/16 ) + 49/8 = -2( y + 3/4 )² + 49/8

\(-2\left(y+\frac{3}{4}\right)^2\le0\forall y\Rightarrow-2\left(y+\frac{3}{4}\right)^2+\frac{49}{8}\le\frac{49}{8}\)

Dấu " = " xảy ra <=> y + 3/4 = 0 => y = -3/4

Vậy N_Max = 49/8 , đạt được khi y = -3/4

c) P = ( 2 -x )( x + 4 ) = -x² - 2x + 8 = -x² - 2x - 1 + 9 = -( x² + 2x + 1 ) + 9 = -( x + 1 )² + 9

\(-\left(x+1\right)^2\le0\forall x\Rightarrow-\left(x+1\right)^2+9\le9\)

Dấu " = " xảy ra <=> x + 1 = 0 => x = -1

Vậy P_Max = 9 , đạt được khi x = -1

M=-x²-4x+2

   =-x²-4x-4+6

   =-(x²+4x+4)+6

   =-(x+2)²=<0 với mọi x

   =>-(x+2)²+6=<6 với mọi x

Dấu "=" xảy ra khi -(x+2)²=0

                             =>x=-2

Vậy....

\(\frac{9^{14}.225^5.8^7}{18^{12}.625^3.24^3}=\frac{\left(3^2\right)^{14}.\left(3^2.5^2\right)^5.\left(2^3\right)^7}{\left(3^2.2\right)^{12}.\left(5^4\right)^3.\left(3.2^3\right)^3}=\frac{3^{28}.3^{10}.5^{10}.2^{21}}{3^{24}.2^{12}.5^{12}.3^3.2^9}=\frac{3^{38}.5^{10}.2^{21}}{3^{27}.2^{21}.5^{12}}=\frac{3^{11}}{5^2}\)

a) \(ĐKXĐ:\) \(x\ne1,x>0\)

\(P=1:\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\right)\)

\(=1:\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)

\(=1:\left[\frac{x+2+x-1-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]\)

\(=1:\frac{\sqrt{x}.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

Vậy \(P=\frac{x+\sqrt{x}+1}{\sqrt{x}}\left(x\ne1,x>0\right)\)

b) Xét hiệu \(P-3=\frac{x+\sqrt{x}+1}{\sqrt{x}}-3\)

\(=\frac{x+\sqrt{x}+1-3\sqrt{x}}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\) \(\forall x>0,x\ne1\)

Do đó : \(P>3\)

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=> \(2x=7\)

=> \(4x=14\)

=> \(4x+1=15\)

Ta có: 2x+1=8

Mà (2x+1)^2= 4x+1

=) 4x+1= 8^2=64 

Vậy 4x+1= 64, mình làm bừa à.

a) Ta có : \(y=\sqrt{2-m}\left(x+1\right)\)

\(=x\sqrt{2-m}+\sqrt{2-m}\)

Để \(y\) là hàm số bậc nhất \(\Leftrightarrow\sqrt{2-m}\ne0\)

\(\Leftrightarrow m\ne4\)

b) Ta có : \(y=\frac{\sqrt{m-5}}{\sqrt{m+5}}x+\sqrt{2}\)

Để \(y\) là hàm số bậc nhất \(\Leftrightarrow\frac{\sqrt{m-5}}{\sqrt{m+5}}\ne0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{m-5}{m+5}\ne0\\m\ne-5\end{cases}}\) \(\Leftrightarrow m\ne\pm5\)

giúp tớ với

( 2m - 3 )( 3n - 2 ) - ( 3m - 2 )( 2n - 3 )

= 6mn - 4m - 9n + 6 - ( 6mn - 9m - 4n + 6 )

= 6mn - 4m - 9n + 6 - 6mn + 9m + 4n - 6

= 5m - 5n

= 5( m - n ) \(⋮\)5 với mọi m, n thuộc Z ( đpcm )