a)x=2015

ai hok biết, giải ra giùm

a) \(\frac{x-15}{2000}+\frac{x-14}{2001}+\frac{x-13}{2002}=\frac{x-12}{2003}+2\)

\(\Rightarrow\left(\frac{x-15}{2000}-1\right)+\left(\frac{x-14}{2001}-1\right)+\left(\frac{x-13}{2002}-1\right)=\left(\frac{x-12}{2003}-1\right)\)

\(\Leftrightarrow\frac{x-2015}{2000}+\frac{x-2015}{2001}+\frac{x-2015}{2002}=\frac{x-2015}{2003}\)

\(\Leftrightarrow\frac{x-2015}{2000}+\frac{x-2015}{2001}+\frac{x-2015}{2002}-\frac{x-2015}{2003}=0\)

\(\Leftrightarrow\left(x-2015\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}\right)=0\)

\(\Rightarrow x-2015=0\)( vì \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}\ne0\))

\(\Leftrightarrow x=2015\)

Vậy tập hợp nghiệm \(S=\left\{2015\right\}\)

\(\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15}.\)

\(\Rightarrow\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15}=\frac{x+11+y+12+z+13}{13+14+15}=\frac{\left(x+y+z\right)+\left(11+12+13\right)}{42}\)

\(=\frac{6+36}{42}=\frac{42}{42}=1\)  ( Áp dụng tính chất dãy tỉ số bằng nhau )

\(\Rightarrow\hept{\begin{cases}\frac{x+11}{13}=1\\\frac{y+12}{14}=1\\\frac{z+13}{15}=1\end{cases}}\Rightarrow\hept{\begin{cases}x+11=13\\y+12=14\\z+13=15\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=2\\z=2\end{cases}}\)

Vậy \(x=y=z=2\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15}=\frac{x+11+y+12+z+13}{13+14+15}\)

\(=\frac{\left(x+y+z\right)+\left(11+12+13\right)}{13+14+15}=\frac{16+36}{42}=\frac{42}{42}=1\)

\(\Rightarrow\frac{x+11}{13}=1\Rightarrow x+11=13\Rightarrow x=13-11=2\)

\(\Rightarrow\frac{y+12}{14}=1\Rightarrow y+12=14\Rightarrow y=14-12=2\)

\(\Rightarrow\frac{z+13}{15}=1\Rightarrow z+13=15\Rightarrow z=15-13=2\)

Vậy \(x=y=z=2\)

Áp dụng tỉ dãy số bằng nhau. Ta có:

\(\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15}\Leftrightarrow\frac{11+12+13+x+y+z}{13+14+15}=6\)

Đặt \(x+y+z=a^3\)

\(\Rightarrow PT=\frac{x+y+z+11+12+13}{13+14+15}=\frac{a^3+11+12+13}{13+14+15}=6\)

\(\Rightarrow\)x ; y  và z \(=6:a^3=6:3=2\)

Vậy dấu = xảy ra khi x = y = z = 6

a. \(\frac{x-15}{2000}+\frac{x-14}{2001}+\frac{x-13}{2003}=\frac{x-12}{2003}+2\)

\(\rightarrow\frac{x}{2000}-\frac{15}{2000}+\frac{x}{2001}-\frac{14}{2001}+\frac{x}{2003}-\frac{13}{2003}=\frac{x}{2003}-\frac{12}{2003}+2\)

\(\rightarrow x.\left(\frac{1}{2000}+\frac{1}{2001}\right)=\frac{15}{2000}+\frac{14}{2001}+\frac{13}{2003}-\frac{12}{2003}+2\)

\(\rightarrow x=2015,5\)

b. \(\left(x^2-6x+11\right)\left(y^2+2y+4\right)=2+4z-z^2\)

\(\rightarrow\left\{{}\begin{matrix}x^2-6x+11=\left(x-3\right)^2+2\ge2\\y^2+2y+4=\left(y+1\right)^2+3\ge3\\2+4z-z^2=-\left(z-2\right)^2+6\le6\end{matrix}\right.\)

\(\rightarrow\left(x^2-6x+11\right)\left(y^2+2y+4\right)\ge6\)

\(\rightarrow\left(x^2-6x+11\right)\left(y^2+2y+4\right)=2+4z-z^2\)

\(\rightarrow\left\{{}\begin{matrix}x=3\\y=-1\\z=2\end{matrix}\right.\)

1 

Ez lắm =)

Bài 1:

Với mọi gt \(x,y\in Q\) ta luôn có: 

\(x\le\left|x\right|\) và \(-x\le\left|x\right|\) 

\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)

Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)

Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)

Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)

Dấu "=" xảy ra khi: \(xy\ge0\)

Bài 3: 

Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) (vì a + b + c = 1)

Do đó: \(\left(x+y+z\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (vì a² + b² + c² = 1)

Vậy: (x + y + z)² = x² + y² + z²

Điều kiện xác định : \(x,y,z\ge0\)

Đặt \(a=\sqrt{x}-13\) , \(b=\sqrt{y}-14\) , \(c=\sqrt{z}-15\)

Ta có hệ : \(\hept{\begin{cases}ab=2\\bc=6\\ac=3\end{cases}}\). Nhân các pt theo vế : \(\left(abc\right)^2=36\Leftrightarrow\orbr{\begin{cases}abc=6\\abc=-6\end{cases}}\)

TH1. Nếu abc = 6 thì kết hợp với mỗi pt ta được : \(\hept{\begin{cases}c=3\\b=2\\a=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=196\\y=256\\z=324\end{cases}}\)

TH2. Nếu \(abc=-6\) thì tương tự ta được \(\hept{\begin{cases}a=-1\\b=-2\\c=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=144\\y=144\\z=144\end{cases}}\)

Vậy ................................................

CHIU THOI

K NHA @@@@@@@ Nguyễn Phúc Lộc 

Theo đầu bài ta có:
\(\hept{\begin{cases}\left(\sqrt{x}-13\right)\left(\sqrt{y}-14\right)=2\\\left(\sqrt{y}-14\right)\left(\sqrt{z}-15\right)=6\\\left(\sqrt{z}-15\right)\left(\sqrt{x}-13\right)=3\end{cases}}\)
\(\Rightarrow\left(\sqrt{x}-13\right)\left(\sqrt{y}-14\right)\cdot\left(\sqrt{y}-14\right)\left(\sqrt{z}-15\right)\cdot\left(\sqrt{z}-15\right)\left(\sqrt{x}-13\right)=2\cdot6\cdot3\)
\(\Rightarrow\left(\sqrt{x}-13\right)\left(\sqrt{y}-14\right)\left(\sqrt{z}-15\right)\cdot\left(\sqrt{x}-13\right)\left(\sqrt{y}-14\right)\left(\sqrt{z}-15\right)=36\)
\(\Rightarrow\left[\left(\sqrt{x}-13\right)\left(\sqrt{y}-14\right)\left(\sqrt{z}-15\right)\right]^2=36\)
\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-13\right)\left(\sqrt{y}-14\right)\left(\sqrt{z}-15\right)=6\\\left(\sqrt{x}-13\right)\left(\sqrt{y}-14\right)\left(\sqrt{z}-15\right)=-6\end{cases}}\)
Từ đây ta xảy ra 2 trường hợp
TH1: Nếu \(\left(\sqrt{x}-13\right)\left(\sqrt{y}-14\right)\left(\sqrt{z}-15\right)=6\) thì:
\(\sqrt{x}-13=\frac{\left(\sqrt{x}-13\right)\left(\sqrt{y}-14\right)\left(\sqrt{z}-15\right)}{\left(\sqrt{y}-14\right)\left(\sqrt{z}-15\right)}=\frac{6}{6}=1\)
\(\Rightarrow\sqrt{x}=14\)
\(\Rightarrow x=196\)

\(\sqrt{y}-14=\frac{\left(\sqrt{x}-13\right)\left(\sqrt{y}-14\right)\left(\sqrt{z}-15\right)}{\left(\sqrt{x}-13\right)\left(\sqrt{z}-15\right)}=\frac{6}{3}=2\)
\(\Rightarrow\sqrt{y}=16\)
\(\Rightarrow y=256\)

\(\sqrt{z}-15=\frac{\left(\sqrt{x}-13\right)\left(\sqrt{y}-14\right)\left(\sqrt{z}-15\right)}{\left(\sqrt{x}-13\right)\left(\sqrt{y}-14\right)}=\frac{6}{2}=3\)
\(\Rightarrow\sqrt{z}=18\)
\(\Rightarrow z=324\)

\(\Rightarrow\hept{\begin{cases}x=196\\y=256\\z=324\end{cases}}\)

7) 5x=4y ⇒\(\dfrac{x}{4}=\dfrac{y}{5}\)

Nhân cả hai vế với \(\dfrac{x}{4}\), ta có: \(\left(\dfrac{x}{4}\right)^2=\dfrac{x}{4}.\dfrac{y}{5}=\dfrac{xy}{20}=\dfrac{20}{20}=1\)

\(\left(\dfrac{x}{4}\right)^2=1\Rightarrow\left[{}\begin{matrix}\dfrac{x}{4}=1\\\dfrac{x}{4}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}y=5\\y=-5\end{matrix}\right.\)

4) áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{0,5}=\dfrac{y}{0,3}=\dfrac{z}{0,2}=\dfrac{z-y+x}{0,2-0,3+0,5}=\dfrac{1}{\dfrac{2}{5}}=\dfrac{5}{2}\)

\(\dfrac{x}{0,5}=\dfrac{5}{2}\Rightarrow x=\dfrac{5}{4}\)

\(\dfrac{y}{0,3}=\dfrac{5}{2}\Rightarrow y=\dfrac{3}{4}\)

\(\dfrac{z}{0,2}=\dfrac{5}{2}\Rightarrow z=\dfrac{1}{2}\)

6) áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x+11}{13}=\dfrac{y+12}{14}=\dfrac{z+13}{15}=\dfrac{x+11+y+12+z+13}{13+14+15}=\dfrac{42}{42}=1\)

\(\dfrac{x+11}{13}=1\Rightarrow x=2\)

\(\dfrac{y+12}{13}=1\Rightarrow y=1\)

\(\dfrac{z+13}{15}=1\Rightarrow z=2\)

7) \(5x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{5}=k\)

\(\Rightarrow x=4k,y=5k\)

\(x.y=20\\ \Rightarrow4k.5k=20\\ \Rightarrow20k^2=20\\ \Rightarrow k^2=1\\ \Rightarrow\left[{}\begin{matrix}k=-1\\k=1\end{matrix}\right.\)

\(x=4k\Rightarrow\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)

\(y=5k\Rightarrow\left[{}\begin{matrix}y=-5\\y=5\end{matrix}\right.\)

Vậy \(\left(x,y\right)=\left\{\left(-4;-5\right);\left(4;5\right)\right\}\)

C1:

\(\text{Ta có: }\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15};x+y+z=6\left(1\right)\)

\(\text{Áp dụng tính dãy tỉ số bằng nhau: }\)

\(\Rightarrow\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15}=\frac{\left(x+11\right)+\left(y+12\right)+\left(z+13\right)}{13+14+15}=\frac{x+11+y+12+z+13}{42}=\frac{\left(x+y+z\right)+\left(11+12+13\right)}{42}\left(2\right)\)

\(\text{Thay (1) và (2), ta được: }\)

\(\frac{6+36}{42}=\frac{42}{42}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x+11}{13}=1\\\frac{y+12}{14}=1\\\frac{z+13}{15}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+11=13\\y+12=14\\z+13=15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=2\\z=2\end{matrix}\right.\)

\(\text{Vậy }x=2;y=2;z=2\)

C2:

\(x+y+z=6\left(1\right)\)

\(\text{Ta có: }\frac{x+11}{13}=\frac{y+12}{14}\Rightarrow x+11=\frac{y+12}{14}.13\Rightarrow x=\frac{13.\left(y+12\right)}{14}-11\)

\(\frac{y+12}{14}=\frac{z+13}{15}\Rightarrow z+13=\frac{y+12}{14}.15\Rightarrow z=\frac{15.\left(y+12\right)}{14}-13\)

\(\text{Thay }x=\frac{13.\left(y+12\right)}{14}-11;z=\frac{15.\left(y+12\right)}{14}-13\text{ vào }\left(1\right)\)

\(\Rightarrow\frac{13.\left(y+12\right)}{14}-11+y+\frac{15.\left(y+12\right)}{14}-13=6\)

\(\Leftrightarrow\frac{13.\left(y+12\right)}{14}+y+\frac{15.\left(y+12\right)}{14}-\left(11+13\right)=6\)

\(\Leftrightarrow\frac{13.\left(y+12\right)}{14}+\frac{14y}{14}+\frac{15.\left(y+12\right)}{14}-24=6\)

\(\Leftrightarrow\frac{13.\left(y+12\right)+14y+15.\left(y+12\right)}{14}=6+24\)

\(\Leftrightarrow\frac{\left(y+12\right).\left(13+15\right)+14y}{14}=30\)

\(\Leftrightarrow\left(y+12\right).28+14y=30.14\)

\(\Leftrightarrow14.\left[\left(y+12\right).2+y\right]=420\)

\(\Leftrightarrow2y+12.2+y=420:14\)

\(\Leftrightarrow3y+24=30\)

\(\Leftrightarrow3y=30-24\)

\(\Leftrightarrow3y=6\)

\(\Leftrightarrow y=6:3\)

\(\Leftrightarrow y=2\)

\(\text{Khi đó: }x=\frac{13.\left(2+12\right)}{14}-11\left(\text{Do y=2}\right)\)

\(\Leftrightarrow x=\frac{13.14}{14}-11\)

\(\Leftrightarrow x=13-11\)

\(\Leftrightarrow x=2\)

\(\text{Khi đó: }z=\frac{15.\left(2+12\right)}{14}-13\left(\text{Do y=2}\right)\)

\(\Leftrightarrow z=\frac{15.14}{14}-13\)

\(\Leftrightarrow z=15-13\)

\(\Leftrightarrow z=2\)

\(\text{Vậy }x=2;y=2;z=2\)

Theo đề bài :

\(\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{13}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{13}=\frac{x+y+z+36}{40}=\frac{21}{20}\)

=> \(\frac{x+11}{13}=\frac{21}{20}\)

=> \(\frac{y+12}{14}=\frac{21}{20}\)

=> \(\frac{z+13}{13}=\frac{21}{20}\)

Rồi đến đây bạn tự làm nốt đi ha !!!

=> x-y /35 = y-z/15 = z-x /21

Theo tính chất dãy tỉ số bằng nhau ta có:

x-y /35 = y-z/15 = z-x /21 = x-y + y-z + z-x / 35+15+21 = 0

=>x-y =0

   y-z =0

   z-x =0

=>x=y=z

 thay vào đẳng thức cầm c/m ta có 2 vế đều = 0 vì y-x=0 và z-y=0 (do x=y=z)