a)

\(x+\left(x+2\right)+\left(x+4\right)+...+\left(x+98\right)=0\)

\(x+x+2+x+4+...+x+98=0\)

\(50x+\left(98+2\right).\left[\left(98-2\right):2+1\right]:2=0\)

\(50x+100.49:2=0\)

\(50x+49.50=0\)

\(50x=0-49.50\)

\(50x=-2450\)

\(x=-2450:50\)

\(x=-49\)

b)

\(\left(x-5\right)+\left(x-4\right)+\left(x-3\right)+...+\left(x+11\right)+\left(x+12\right)=99\)

\(x+x+x+...+x-5-4-3-...+11+12=99\)

\(18x+6+7\text{+ 8 + 9 + 10 + 11 + 12 = 99}\)

\(18x+63=99\)

\(18x=99-63\)

\(18x=36\)

\(x=36:18\)

\(x=2\)

giúp mình với, mình đang vội!

a) x + (x + 2) + (x + 4) + ... + (x + 98) = 0

x + x + 2 + x + 4 + ... + x + 98 = 0

50x + (98 + 2).[(98 - 2) : 2 + 1]:2 = 0

50x + 100 .49 : 2 = 0

50x + 49.50 = 0

50x = 0 - 49.50

50x = -2450

x = -2450 : 50

x = -49

b) (x - 5) + (x - 4) + (x - 3) + ... + (x + 11) + (x + 12) = 99

x + x + x + ... + x - 5 - 4 - 3 - ... + 11 + 12 = 99

18x + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 99

18x + 63 = 99

18x = 99 - 63

18x = 36

x = 36 : 18

x = 2

-29-9(2x-1)\(^2\)= -110

(=) 9(2x-1)² = (-29) +110

(=) 9(2x-1)² = 81

(=) (2x-1)² =81: 9

(=) (2x-1)² =9

(=) (2x-1)² = 3² =(-3)²

\(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)

\(\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)

\(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

vậy : ........

a,\(-29-9\left(2x-1\right)^2=-110\)

\(=>-29+110=9.\left(2x-1\right)^2\)

\(=>81=9.\left(2x-1\right)^2\)

\(=>\left(2x-1\right)^2=9\)

\(=>\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}=>\orbr{\begin{cases}x=\frac{4}{2}=2\\x=\frac{-2}{2}=-1\end{cases}}}\)

b,\(-2.\left(-x-5\right)+28=20-3.4\)

\(=>2x+10+28=20-12=8\)

\(=>2x=8-28-10=-30\)

\(=>x=\frac{-30}{2}=-15\)

Lời giải:
a. Vì $x,y$ thuộc $Z$ nên $x-3, y+5\in\mathbb{Z}$. Tích của chúng $=11$ nên ta có bảng sau:

b. Vì $x,y\in\mathbb{Z}$ nên $2x+1, 6-y\in\mathbb{Z}$.

Với $x$ nguyên thì $2x+1$ là số nguyên lẻ nên ta có bảng sau:

a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)

b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)

c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)

\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)

a) \(x\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b) \(\left(-7-x\right)\left(-x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)

c) \(\left(x+3\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

d) \(\left(x-3\right)\left(x^2+12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)

\(\Rightarrow x=3\)

e) \(\left(x+1\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow-1\le x\le2\)

f) \(\left(x-3\right)\left(x-5\right)\le0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow3\le x\le5\)

a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3

c) => \(\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

a: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)

=>x=-3; y=28; z=-60

b: 5/12=x/-72

=>x=-72*5/12=-6*5=-30

c: =>x+3=-5

=>x=-8

a, \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-3,2y-6\in Z\\x-3,2y-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\end{matrix}\right.\)

Ta có bảng:

Vậy không có x,y thỏa mãn đề bài 

b, tương tự câu a

 \(c,xy-5x+2y=7\\ \Rightarrow x\left(y-5\right)+2y-10=-3\\ \Rightarrow x\left(y-5\right)+2\left(y-5\right)=-3\\ \Rightarrow\left(x+2\right)\left(y-5\right)=-3\)

Rồi làm tương tự câu a

\(d,xy-3x-4y=5\\ \Rightarrow x\left(y-3\right)-4y+12=17\\ \Rightarrow x\left(y-3\right)-4\left(y-3\right)=17\\ \Rightarrow\left(x-4\right)\left(y-3\right)=17\)

Rồi làm tương tự câu a

a: 2x-3y-4z=24

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)

=>x=-6/7; y=-36/7; z=-18/7

b: 6x=10y=15z

=>x/10=y/6=z/4=k

=>x=10k; y=6k; z=4k

x+y-z=90

=>10k+6k-4k=90

=>12k=90

=>k=7,5

=>x=75; y=45; z=30

d: x/4=y/3

=>x/20=y/15

y/5=z/3

=>y/15=z/9

=>x/20=y/15=z/9

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)

=>x=500; y=375; z=225

a) (x-1):2/3=-2/5

=>x-1=-4/15

=>x=11/15

b) |x-1/2|-1/3=0

=>|x-1/2|=1/3

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\) 

c) Tương Tự câu B