ĐK : x >= 0 ; x khác 4

\(F=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{4\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+2\sqrt{x}-4\sqrt{x}+8-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}}{\sqrt{x}+2}\)

muốn nhanh phải dùng máy tính

kq=8999600

x càng lớn thì biểu thức càng lớn nên cái này k có max bạn nhé 

ok cảm ơn bạn

Ta có: 5^2n-25=25^n-25

=(...25)-25=(...00) chia hết cho 100 nha

5^2n - 25 : 100

= 25^n - 25 

= (.....25)-25 =(.....00) chia hết cho 100 nha bạn 

chúc bạn học tốt ạ 

Hạ \(DH\perp AB,CK\perp AB\).

Tam giác \(AHD\)vuông tại \(H\)có\(\widehat{ADH}=30^o\)nên \(AH=\frac{1}{2}AD=1\left(cm\right)\)

\(HD=\sqrt{AD^2-AH^2}=\sqrt{2^2-1^2}=\sqrt{3}\left(cm\right)\)

Tương tự \(BK=1\left(cm\right)\).

\(DC=HK=AB-AH-BK=2,5\left(cm\right)\)

\(S_{ABCD}=\frac{AB+CD}{2}.DH=\frac{4,5+2,5}{2}.\sqrt{3}=\frac{7\sqrt{3}}{2}\left(cm^2\right)\)

\(\sqrt{a}+\sqrt{b}=m\Leftrightarrow m-\sqrt{a}=\sqrt{b}\Rightarrow m^2-2m\sqrt{a}+a=b\)

\(\Leftrightarrow\sqrt{a}=\frac{m^2+a-b}{2m}\)là số hữu tỉ. 

Tương tự cũng suy ra \(\sqrt{b}\)là số hữu tỉ. 

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

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

\(< =>\left(x^2-1\right)\left(x^2+4\right)=0\)

\(< =>\left(x-1\right)\left(x+1\right)=0\)(vì x² + 4 > 0)

\(< =>\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Trả lời:

Câu 1: 

a, \(\left(5x+1\right)\left(2x-3\right)-\left(7x-2\right)\left(x+3\right)+5\)

\(=10x^2-15x+2x-3-\left(7x^2+21x-2x-6\right)+5\)

\(=10x^2-13x-3-7x^2-21x+2x+6+5\)

\(=3x^2-32x+8\)

b, \(\left(3x-2\right)^2+\left(2x+5\right)^2-\left(3x+4\right)\left(3x-4\right)\)

\(=9x^2-12x+4+4x^2+20x+25-9x^2+16\)

\(=4x^2+8x+45\)

c, \(\left(3x+2\right)\left(9x^2-6x+4\right)+\left(5-3x\right)\left(25+15x+9x^2\right)-9\)

\(=27x^3+8+125-27x^3-9\)

\(=124\)

d, \(\left(x-3\right)^3-\left(x+2\right)\left(x-2\right)\left(x-1\right)\)

\(=x^3-9x^2+27x-27-\left(x^2-4\right)\left(x-1\right)\)

\(=x^3-9x^2+27x-27-\left(x^3-x^2-4x+4\right)\)

\(=x^3-9x^2+27x-27-x^3+x^2+4x-4\)

\(=-8x^2+31x-31\)

e, \(\left(x+2y\right)^3-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=x^3+6x^2y+12xy^2+8y^3-\left(8x^3-y^3\right)\)

\(=x^3+6x^2y+12xy^2+8y^3-8x^3+y^3\)

\(=-7x^3+9y^3+6x^2y+12xy^2\)

f, \(\left(x-3y\right)^3-\left(2x-y\right)\left(3x+2y\right)\left(3x-2y\right)\)

\(=x^3-9x^2y+27xy^2-\left(2x-y\right)\left(9x^2-4y^2\right)\)

\(=x^3-9x^2y+27xy^2-27y^3-\left(18x^3-8xy^2-9x^2y+4y^3\right)\)

\(=x^3-9x^2y+27xy^2-27y^3-18x^3+8xy^2+9x^2y-4y^3\)

\(=-17x^3-31y^3+35xy^2\)

Câu 2 : a ) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow2x^2-10x-3x-2x^2=-13x=26\Leftrightarrow x=-2\)

b) \(\left(3x+1\right)^2+\left(2x-5\right)^2=13\left(x+2\right)\left(x-2\right)\)

\(\Leftrightarrow9x^2+6x+1+4x^2-20x+25=13\left(x^2-4\right)\)

\(\Leftrightarrow13x^2-14x+26=13x^2-52\Leftrightarrow13x^2-14x+26-13x^2+52=0\)

\(\Leftrightarrow78-14x=0\Leftrightarrow78=14x\Leftrightarrow x=\frac{78}{14}=\frac{39}{7}\)