\(ĐKXĐ:m\ge0;m\ne1\)

\(a,M=\frac{\sqrt{m}-1}{\sqrt{m}+1}+\frac{\sqrt{m}+1}{\sqrt{m}-1}\)

\(=\frac{\left(\sqrt{m}-1\right)^2+\left(\sqrt{m}+1\right)^2}{\left(\sqrt{m}+1\right)\left(\sqrt{m}-1\right)}\)

\(=\frac{m-2\sqrt{m}+1+m+2\sqrt{m}+1}{m-1}\)

\(=\frac{2m+2}{m-1}\)

b,Để M nguyên thì \(\frac{2m+2}{m-1}=\frac{2\left(m-1\right)}{m-1}+\frac{4}{m-1}=2+\frac{4}{m-1}\) nguyên

\(\Rightarrow m-1\inƯ\left(4\right)=\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow m=\left\{2;3;5;0;-1;-3\right\}\)

\(KethopDKXD:m=\left\{2;3;5;0\right\}\)

\(\sqrt[3]{\frac{3}{4}}:\sqrt[3]{\frac{9}{16}}=\frac{3}{4}:\frac{9}{16}\)

\(=\frac{3}{4}.\frac{16}{9}=\frac{4}{3}\)

Học tốt!!!!!!!!!!

ban oi ban lam sai oi

làm hộ mình nha

\(PT\Leftrightarrow x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=1\)

\(\Leftrightarrow x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}=1-2x+x^2\)

\(\Leftrightarrow\sqrt{x+\frac{1}{4}}=x^2-3x+\frac{1}{2}\)

\(\Leftrightarrow x+\frac{1}{4}=x^4+9x^2+\frac{1}{4}-6x^3-3x+x^2\)

\(\Leftrightarrow x^4-6x^3+10x^2-4x=0\)

\(\Leftrightarrow x\left(x^3-6x^2+10x-4\right)=0\)

Vì x³-6x²+10x-4 \(\ne\)0

nên x=0

Vậy..................................

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

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

\(=1+2\sqrt{2}+2-3\)

\(=2\sqrt{2}\)

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

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

\(=1+4\sqrt{3}+12-2\)

\(=9+4\sqrt{3}\)

b) \(\sqrt{25a^2}+3a\) \(=5\left|a\right|+3a\)

Vì a > 0 => |a| = a

=> 5|a| + 3a = 5a + 3a = 8a

cái nịt

\(R-CHO+Ag_2O\rightarrow R-COOH+2Ag\downarrow\)

\(Mg\)                                                                  \(2.108\)

\(2,2\)                                                                    \(10,8\)

\(M=\frac{216.2,2}{10,8}=44\)

\(M=14n+16=44\Rightarrow n=\frac{44-16}{14}=\frac{28}{14}=2\)

\(C_2HO_4\Rightarrow CH_3-CHO\)

ai trả lời được mình cho 1 k

Để PT có nghiệm phân biệt thì: \(\Delta^'>0\)

Hay: \(\left[-\left(m+1\right)\right]^2-\left(m^2-10\right)>0\)

\(\Leftrightarrow m^2+2m+1-m^2+10>0\)

\(\Leftrightarrow2m>-11\)

\(\Leftrightarrow m>-\frac{11}{2}\)

Theo Vi-et, ta có: \(\hept{\begin{cases}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2-10\end{cases}}\) (1)

Ta có: \(C=x_1^2+x_2^2=x_1^2+2x_1x_2+x^2_2-2x_1x_2=\left(x_1+x_2\right)^2-2x_1x_2\)

Thay (1) vào C, ta được:

\(C=4\left(m+1\right)^2-2\left(m^2-10\right)\)

\(=4m^2+8m+4-2m^2+20\)

\(=2m^2+8m+24\)

\(=2\left(m^2+4m+12\right)\)

\(=2\left(m^2+4m+4+8\right)\)

\(=2\left(m+2\right)^2+16\ge16\forall m\)

=> Min C = 16 tại m = - 2 (tm)

=.= hk tốt!!