Ta có:

\(4x^2+12x+100=\left(2x+3\right)^2+91\)

\(\Rightarrow B=\frac{-9}{\left(2x+3\right)^2+91}\)

Vì \(\left(2x+3\right)^2\ge0;\forall x\)

\(\Rightarrow\left(2x+3\right)^2+91\ge0+91;\forall x\)

\(\Rightarrow\frac{9}{\left(2x+3\right)^2+91}\le\frac{9}{91};\forall x\)

\(\Rightarrow\frac{-9}{\left(2x+3\right)^2+91}\ge\frac{-9}{91};\forall x\)

Dấu '"=" xảy ra \(\Leftrightarrow2x+3=0\)

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

Vậy MIN \(B=\frac{-9}{91}\)\(\Leftrightarrow x=\frac{-3}{2}\)

TL:

\(B=\frac{-9}{\left(2x+6\right)^2+64}\) 

 Để B_min \(\Rightarrow\left(2x+6\right)^2+64\) nhỏ nhất

Mà \(\left(2x+6\right)^2+64\ge64\forall x\in R\) 

dấu "=" xảy ra <=> \(\left(2x+6\right)^2=0\Leftrightarrow2x+6=0\Leftrightarrow2x=-6\Leftrightarrow x=-3\) 

=>B_min =\(\frac{-9}{64}\) tại x=-3

Vậy.......

a, b sai đề nhé , sửa lại :

\(a,x^7+x^5+1=x^7+x^6+x^5-x^6+1=....\)

\(b,x^5+x+1=x^5-x^2+x^2+x+1=....\)

\(c,x^{11}+x+1=x^{11}-x^8+x^8-x^5+x^5-x^2+x^2+x+1=...\)

\(d,x^8+x^7+1=x^8+x^7+x^6-x^6+1=...\)

\(e,x^5+x^4+2x^2-1\)

Câu e tớ chịu , các câu trên tớ chỉ cho cậu hướng tách các hạng tử thôi, để cậu dễ dàng nhóm các nhân tử chung là \(x^2+x+1\), câu nào chưa làm được nữa thì để tớ giải rõ hơn nha

Ta có:

A = \(\frac{-5}{3x^2-6x+108}=\frac{-5}{3\left(x^2-2x+1\right)+105}=\frac{-5}{3\left(x-1\right)^2+105}\)

Ta luôn có: (x - 1)² \(\ge\)0 \(\forall\)x ---> 3(x - 1)² \(\ge\)0 \(\forall\)x

=> 3(x - 1)² + 105 \(\ge\) 105 \(\forall\)x

=> \(-\frac{5}{3\left(x-1\right)^2+105}\ge-\frac{1}{21}\)\(\forall\)x

hay A \(\ge\)-1/21 \(\forall\)x

Dấu "=" xảy ra khi: x - 1 = 0 <=> x = 1

Vậy A_min = -1/21 tại x = 1

Ta có:

\(A=\frac{-5}{3x^2-6x+108}=\frac{-5}{3x^2-6x+3+105}=\frac{-5}{3\left(x-1\right)^2+105}\)

\(3\left(x-1\right)^2+105\ge105\)\(,\forall x\)

\(\Rightarrow\frac{1}{3\left(x-1\right)^2+105}\le\frac{1}{105}\Rightarrow\frac{-1}{3\left(x-1\right)^2+105}\ge-\frac{1}{105}\)\(\Rightarrow\frac{-5}{3\left(x-1\right)^2+105}\ge-\frac{5}{105}=-\frac{1}{21}\) \(GTNNA=-\frac{1}{21}\Leftrightarrow3\left(x-1\right)^2=0\)

                                         \(\Rightarrow\left(x-1\right)^2=0\)

                                                 \(\Rightarrow x-1=0\)

                                                                \(x=1\)

Vậy \(GTNNA=-\frac{1}{21}\Leftrightarrow x=1\)

2X.X - 2/7X = 6

X.X(2 - 2/7) = 6

X.X.12/7 = 6

X.X = 6 : 12/7

X.X = 6.7/12

X.X = 42/12

X THUỘC TẬP HỢP RỖNG NHA BẠN => Ko có x phù hợp

1. \(\frac{x+1}{x+5}=\frac{x+3}{x+2}\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=\left(x+3\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x+1\right)x+\left(x+1\right).2=\left(x+3\right)x+\left(x+3\right).5\)

\(\Leftrightarrow x^2+x+2x+2=x^2+3x+5x+15\)

\(\Leftrightarrow x^2+3x+2=x^2+8x+15\)

\(\Leftrightarrow x^2+3x-x^2-8x=15-2\)

\(\Leftrightarrow-5x=13\)

\(\Leftrightarrow x=\frac{-13}{5}\)

Vậy ...

Cảm ơn bn nha 

x+x+1+x+2+x+3+...+x+10=2029099

<=>11x+45=2029099

<=>11x=2029054

<=>x=184459.4545

chac v . hc tot

x + (x + 1) + (x + 2) + (x + 3) + ... + (x + 10) = 2029099

100x + (1+2+...+10) = 2029099

100x + 55 = 2029099

100x = 2029099 - 55

100x = 2029044

x = 2029099 : 100

x = 20290,99

\(\frac{x+\sqrt{xy}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}.\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)^2}{x-y}\)

\(a,\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)

\(=\sqrt{13+6\sqrt{4+\sqrt{1-2.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}}\)

\(=\sqrt{13+6\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}}\)

\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)

\(=\sqrt{13+6\sqrt{3+2\sqrt{2}}}\)

\(=\sqrt{13+6\sqrt{1+2\sqrt{2}+2}}\)

\(=\sqrt{13+6\sqrt{\left(1+\sqrt{2}\right)^2}}\)

\(=\sqrt{13+6\left(1+\sqrt{2}\right)}=\sqrt{13+6+\sqrt{12}}\)

\(=\sqrt{19+2\sqrt{3}}\)

a) = \(\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)

=    \(\sqrt{13+6\sqrt{4+\sqrt{8-2.2\sqrt{2}+1}}}\)

=    \(\sqrt{13+6\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}}\)

=    \(\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)

=     \(\sqrt{13+6\sqrt{2+2\sqrt{2}+1}}\)

=     \(\sqrt{13+6\left(\sqrt{2}+1\right)}\)

=     \(\sqrt{13+6\sqrt{2}+6}=\sqrt{19+6\sqrt{2}}\)

=      \(\sqrt{18+2.3\sqrt{2}+1}\)

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

=       \(3\sqrt{2}+1\)

2¹⁹³⁰ - 9¹⁹⁴⁵

=...4-..9

=...14-..9

=....5

chắc vậy 

\(2^{1930}.9^{1945}\)

\(=2^{4.482+2}.9^{4.486+1}\)

\(=2^{4.482}.2^2.9^{4.486}.9\)

\(=...6.4....1.9\)

\(=...6\)

Chữ số tận cùng là 6.