o ai biet lam au nay thi giup minh nhe

co ai giup minh voi

1) tự làm (thực hiện từ dưới lên)

2) B = \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{4}\right)^5.3}{\frac{\frac{1}{1024}.1}{3}-\left(\frac{1}{2}\right)^{11}}\)

      = \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{2}\right)^{10}.3}{\left(\frac{1}{2}\right)^{10}.\frac{1}{3}-\left(\frac{1}{2}\right)^{10}.\frac{1}{2}}\)

     = \(\frac{\left(\frac{1}{2}\right)^{10}.\left(5-3\right)}{\left(\frac{1}{2}\right)^{10}.\left(\frac{1}{3}-\frac{1}{2}\right)}\)

     = \(\frac{2}{-\frac{1}{6}}\)= 2 . (-6) = -12

1) \(5+\frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{4}}}}=5+\frac{15}{7}=\frac{5}{1}+\frac{15}{7}=\frac{50}{7}\)

Ta có: \(A=\frac{\sqrt{x}-3}{\sqrt{x}+2}=\frac{\sqrt{x}+2-5}{\sqrt{x}+2}=1-\frac{5}{\sqrt{x}+2}=-1\)

a)Thay x = 1/4 vào A,ta có \(A=1-\frac{5}{\sqrt{x}+2}=1-\frac{5}{\sqrt{\frac{1}{4}}+2}=-1\)

b) Theo kết quả câu a) khi x = 1/4  thì A = -1

Vậy x = 1/4

c)Để A nhận giá trị nguyên thì \(\frac{5}{\sqrt{x}+2}\) nguyên.

Hay \(\sqrt{x}+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Đến đây bí.

tu ve hinh : 

xet tamgiac ABI va tamgiac ACI co :

AB = AC => tamgiac ABC can tai A => AB = AC va goc ABC = goc ACB (dn)

BI = IC do I la trung diem cua BC 

=>  tamgiac ABI = tamgiac ACI (c - g - c)        (1)

=> goc BAI = goc CAI (dn) ma` AI nam giua AB va AC

=> AI la phangiac cua goc BAC (dn)

b, 

goc ABM + goc ABI = 180 (kb)

goc ACN + goc ACI = 180 (kb)

goc ABI = goc ACI (cau a)

=> goc MBA = goc ACN 

xet tamgiac ABM va tamgiac ACN co : AB = AC (gt)

MB = CN (gt)

=> tamgiac MBA = tamgiac NCA (c - g - c)

=> AM = AN (dn)

c,  (1) => goc AIB = goc AIC  

ma` goc AIB + goc AIC = 180 (kb)

=> goc AIB = 90

=> AI | BC (dn)

Ta có : \(\frac{2}{3}x=\frac{3}{4}y=\frac{5}{6}z\)

\(\Rightarrow\frac{8}{12}x=\frac{9}{12}y=\frac{10}{12}z\)

\(\Rightarrow8x=9y=10z\)

\(\Rightarrow\frac{8x}{360}=\frac{9y}{360}=\frac{10z}{360}\)

\(\Rightarrow\frac{x}{45}=\frac{y}{40}=\frac{z}{36}\)

\(\Rightarrow\frac{x}{45}=\frac{2y}{80}=\frac{z}{36}\)

ADTCDTSBN , ta có 

\(\frac{x}{45}=\frac{2y}{80}=\frac{z}{36}=\frac{x+2y+z}{45+80+36}=\frac{-39}{161}\)

Do đó : \(x=\frac{-39}{161}.45=-10\frac{145}{161}\)

\(y=\frac{-39}{161}.40=-9\frac{111}{161}\)

\(z=\frac{-39}{161}.36=-8\frac{116}{161}\)

Vậy x , y , z lần lượt là .........

\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{x}\left(1+2+3+...+x\right)\)

\(B=1+\frac{1}{2}\left(1+2\right)\cdot2:2+\frac{1}{3}\left(1+3\right)\cdot3:2+...+\frac{1}{x}\left(1+x\right)\cdot x:2\)

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

\(B=1+\frac{\left(1+1+...+1\right)+\left(2+3+...+x\right)}{2}\)

De B = 115

=> \(\frac{\left(1+1+...+1\right)+\left(2+3+...+x\right)}{2}=114\)

=> (1 + 1 + ... + 1) + (2 + 3 + ... + x) = 228

den day chju :v

\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.............+\frac{1}{x}\left(1+2+3+............+x\right)\)

\(=1+\frac{1}{2}\frac{2.3}{2}+\frac{1}{3}\frac{3.4}{2}+...........+\frac{1}{x}\frac{x\left(x+1\right)}{2}\)

\(=\frac{1}{2}\left(2+3+4+.............+\left(x+1\right)\right)\)

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

\(=\frac{1}{4}\left(x+3\right)x\)

\(B=115\Leftrightarrow\frac{1}{4}.x\left(x+3\right)=115\)

\(\Leftrightarrow x\left(x+3\right)=115.4\)

\(\Leftrightarrow x\left(x+3\right)=20.23\)

\(\Leftrightarrow x=20\)

\(\text{Vì }a,b,c\inℕ^∗\Rightarrow\hept{\begin{cases}\frac{a}{a+b}>\frac{a}{a+b+c}\\\frac{b}{b+c}>\frac{b}{a+b+c}\\\frac{c}{c+a}>\frac{c}{a+b+c}\end{cases}\Rightarrow M>\frac{a+b+c}{a+b+c}=1}\)(1)

\(\hept{\begin{cases}\frac{a}{a+b}< \frac{a+c}{a+b+c}\\\frac{b}{b+c}< \frac{b+a}{a+b+c}\\\frac{c}{a+c}< \frac{c+b}{c+a+b}\end{cases}}\Rightarrow M< \frac{2.\left(a+b+c\right)}{a+b+c}=2\)(2) (chỉ áp dụng cho p/s có tử  bé hơn mẫu)

từ (1) và (2) => 1<M<2 => M không phải là STN

\(M=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)

Ta có

\(\frac{a}{a+b+c}< \frac{a}{a+b}< \frac{a+c}{a+b+c}\)

\(\frac{b}{b+c+a}< \frac{b}{b+c}< \frac{b+a}{b+c+a}\)

\(\frac{ c}{c+a+b}< \frac{c}{c+a}< \frac{c+b}{c+a+b}\)

\(\Rightarrow1< M< 2\Rightarrow\)M không phải là số tự nhiên

\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+99\right|=100x\)

\(\left|x+1\right|\ge0;\left|x+2\right|\ge0;...;\left|x+99\right|\ge0\)

\(\Rightarrow100x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+1+x+2+x+3+...+x+99=100x\)

\(\Rightarrow99x+1+2+3+...+99=100x\)

\(\Rightarrow99x+4950=100x\)

\(\Rightarrow-x=-4950\)

\(\Rightarrow x=4950\)

\(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{49\cdot50}\right|=50x\)

\(\left|x+\frac{1}{1\cdot2}\right|\ge0;\left|x+\frac{1}{2\cdot3}\right|\ge0;...;\left|x+\frac{1}{49\cdot50}\right|\ge0\)

\(\Rightarrow50x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{49\cdot50}\)

\(\Rightarrow49x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=50x\)

\(\Rightarrow49x+\frac{49}{50}=50x\)

tu lam 

\(a;\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+..............+\left|x+99\right|=100x^{\left(1\right)}\)

Ta có \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+3\right|\ge0;.............;\left|x+99\right|\ge0\)

\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow100x\ge0\Rightarrow x\ge0\)

Với \(x\ge0\).Từ (1) \(\Rightarrow x+1+x+2+x+3+..................+x+99=100x\)

\(\Rightarrow\left(x+x+x+........+x\right)+\left(1+2+3+..........+99\right)=100x\)

\(\Rightarrow99x+4950=100x\)

\(\Rightarrow x=4950\)(t/m đk x > =  0)

\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+.........+\left|x+\frac{1}{49.50}\right|=50x^{(∗)}\)

\(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;............;\left|x+\frac{1}{49.50}\right|\ge0\)

\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow50x\ge0\Rightarrow x\ge0\)

Với x > = 0 .Từ (*) \(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+............+x+\frac{1}{49.50}=50x\)

\(\Rightarrow\left(x+x+x+.......+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{49.50}\right)=50x\)

\(\Rightarrow49x+\left(1-\frac{1}{50}\right)=50x\)

\(\Rightarrow49x+\frac{49}{50}=50x\)

\(\Rightarrow x=\frac{49}{50}\)(t/m đk \(x\ge0\))

Trước hết ta chứng minh bổ đề: \(|a|+|b|\ge|a+b|.\left(1\right)\)

CM: \(\left(1\right)\Leftrightarrow\left(|a|+|b|\right)^2\ge\left(|a+b\right)^2\)

                  \(\Leftrightarrow a^2+b^2+2|ab|\ge a^2+b^2+2ab\)

                  \(\Leftrightarrow2|ab|\ge2ab\)

                  \(\Leftrightarrow\left|ab\right|\ge ab\)(điều này đúng do tính chất của giá trị tuyệt đối).

Vậy ta có đpcm. Dấu bằng xảy ra \(\Leftrightarrow ab\ge0.\)

a) A = \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|+\left|x-2\right|.\)

Ta thấy rằng \(\left|x-2\right|\ge0\)với mọi x.

Áp dụng bổ đề trên ta có:

\(A\ge\left|x-1+3-x\right|+0=\left|2\right|+0=2+0=2.\)

Dấu bằng xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)\left(3-x\right)\ge0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}1\le x\le3\\x=2\end{cases}}\Leftrightarrow x=2.\)

Vậy GTNN của A bằng 2 khi x = 2.

b) Áp dụng bổ đề trên ta có:\(B=\left|x-4\right|+\left|7-x\right|+\left|x-5\right|+\left|6-x\right|\ge\left|x-4+7-x\right|+\left|x-5+6-x\right|=\left|3\right|+\left|1\right|=3+1=4.\)

Dấu bằng xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-4\right)\left(7-x\right)\ge0\\\left(x-5\right)\left(6-x\right)\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}4\le x\le7\\5\le x\le6\end{cases}\Leftrightarrow}5\le x\le6}\)(vì với mọi x nằm giữa 5 và 6 thì cũng nằm giữa 4 và 7).

Vậy GTNN của B bằng 4 khi \(5\le x\le6.\)

a;\(A=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)

\(\Rightarrow A=\left|x-1\right|+\left|x-2\right|+\left|3-x\right|\)

Ta có +) \(\left|x+1\right|+\left|3-x\right|\ge\left|x+1+3-x\right|=4\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)\left(3-x\right)\ge0\Leftrightarrow1\le x\le3\)

+)\(\left|x-2\right|\ge0\)Dấu "=" xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

\(\Rightarrow A=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\ge2\)

\(\Rightarrow A_{min}=2\Leftrightarrow\hept{\begin{cases}1\le x\le3\\x=2\end{cases}\Leftrightarrow x=2}\)

b;\(B=\left|x-4\right|+\left|x-5\right|+\left|x-6\right|+\left|x-7\right|\)

\(\Rightarrow B=\left|x-4\right|+\left|x-5\right|+\left|6-x\right|+\left|7-x\right|\)

Ta có +) \(\left|x-4\right|+\left|7-x\right|\ge\left|x-4+7-x\right|=3\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-4\right)\left(7-x\right)\ge0\Leftrightarrow4\le x\le7\)

+) \(\left|x-5\right|+\left|6-x\right|\ge\left|x-5+6-x\right|=1\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-5\right)\left(6-x\right)\ge0\Leftrightarrow5\le x\le6\)

\(\Rightarrow B=\left|x-4\right|+\left|x-5\right|+\left|x-6\right|+\left|x-7\right|\ge4\)

\(\Rightarrow B_{min}=4\Leftrightarrow\hept{\begin{cases}4\le x\le7\\5\le x\le6\end{cases}\Leftrightarrow5\le x\le6}\)