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!!

giúp mình với mấy bạn

p

a) \(ab+bc+ca=1\)\(\Rightarrow\)\(\hept{\begin{cases}a^2b^2+b^2c^2+c^2a^2=1-2abc\left(a+b+c\right)\\\left(a+b+c\right)^2-2=a^2+b^2+c^2\end{cases}}\)

\(A=\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}=\sqrt{a^2b^2c^2+a^2b^2+b^2c^2+c^2a^2+a^2+b^2+c^2+1}\)

\(A=\sqrt{a^2b^2c^2-2abc\left(a+b+c\right)+\left(a+b+c\right)^2}\)

\(A=\sqrt{\left(abc-a-b-c\right)^2}=\left|abc-a-b-c\right|\)

Do a, b, c là các số hữu tỉ nên \(\left|abc-a-b-c\right|\) là số hữu tỉ 

b) \(B=\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2}}}}>\sqrt{1+\sqrt{1+\sqrt{1+...+\sqrt{1}}}}=1\)

\(B< \sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{4}}}}=\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2+2}}}}=\sqrt{2+2}=2\)

=> \(1< B< 2\) B không là số tự nhiên 

c) câu này có ng làm r ib mk gửi link 

à chỗ câu b) mình nhầm tí nhé 

\(B=\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2}}}}>\sqrt{1+\sqrt{1+\sqrt{1+...+\sqrt{1}}}}>1\)

Sửa dấu "=" thành ">" hộ mình 

\(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}\)\(-\sqrt{60}\)

\(=2\sqrt{3}.\sqrt{3}\)\(+\sqrt{5}.\sqrt{3}\)\(-\sqrt{4.15}\)

\(=2.3+\sqrt{15}-2\sqrt{15}\)

\(=6+\sqrt{15}.\left(1-2\right)\)

\(=6-\sqrt{15}\)

\(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}-\sqrt{60}\)

\(=2\sqrt{3}.\sqrt{3}+\sqrt{5}.\sqrt{3}-\sqrt{60}\)

\(=2.3+\sqrt{5.3}-\sqrt{60}\)

\(=6+\sqrt{15}-\sqrt{60}\)

\(=6-\sqrt{15}\)

a, \(16x^2-5=0\)

\(\Rightarrow16x^2=5\)

\(\Rightarrow x^2=\frac{5}{16}\)

\(\Rightarrow x=\sqrt{\frac{5}{16}}\Rightarrow x=\frac{\sqrt{5}}{4}\)

b, \(2\sqrt{x-3}=4\)

\(\Rightarrow\sqrt{x-3}=4:2\)

\(\Rightarrow\sqrt{x-3}=2\)

\(\Rightarrow x-3=4\)

\(\Rightarrow x=4+3\)

\(\Rightarrow x=7\)

c, \(\sqrt{4x^2-4x+1}=3\)

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

\(\Rightarrow2x-1=3\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

d, \(\sqrt{x+3}\ge5\)

\(\Rightarrow x+3\ge25\)

\(\Rightarrow x\ge22\)

e, \(\sqrt{3x-1}< 2\)

\(\Rightarrow3x-1< 4\)

\(\Rightarrow3x< 5\)

\(\Rightarrow x< \frac{5}{3}\)

g, \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

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

\(\Rightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

\(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)

\(\Rightarrow\sqrt{x-3}=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

a) \(16x^2-5=0\)

\(\Leftrightarrow16x^2=5\)

\(\Leftrightarrow x^2=\frac{5}{16}\)

\(\Leftrightarrow x=\pm\sqrt{\frac{5}{16}}\)

b) \(2\sqrt{x-3}=4\)

\(\Leftrightarrow\sqrt{x-3}=2\)

\(\Leftrightarrow x-3=4\)

\(\Leftrightarrow x=7\)

c) \(\sqrt{4x^2-4x+1}=3\)

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

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

d) \(\sqrt{x+3}\ge5\)

\(\Leftrightarrow x+3\ge25\)

\(\Leftrightarrow x\ge22\)

e) \(\sqrt{3x-1}< 2\)

\(\Leftrightarrow3x-1< 4\)

\(\Leftrightarrow3x< 5\)

\(\Leftrightarrow x< \frac{5}{3}\)

g) \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

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

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

Vì \(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)

\(\Leftrightarrow\sqrt{x-3}=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Ta có \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\ge xy\left(x+y\right)\)

Áp dụng ta có 

\(a+b\ge\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\)

=> \(a+b+1\ge\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)\)

Khi đó

\(A\le\frac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\sqrt[3]{abc}\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)}=1\)

MaxA=1

Dấu bằng xảy ra khi a=b=c=1

Áp dụng bất đẳng thức cosi schwarz 

 \(A\ge\frac{\left(x+y+z\right)^2}{2x^2+2y^2+2z^2+5\left(xy+yz+xz\right)}=\frac{9}{18+\left(xy+yz+xz\right)}\)

Mà \(xy+yz+xz\le\frac{1}{3}\left(x+y+z\right)^2=3\)

=> \(A\ge\frac{9}{18+3}=\frac{3}{7}\)

MinA=3/7

Dấu bằng xảy ra khi x=y=z=1