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(x+x+x+x+x+...+x)+(1+3+5+...+99)=0

50x + 2500 = 0

50x=0- 2500

50x =-2500

x=-2500:50

x=-50 

Vậy x=-50

a)           (x+1) + (x+3) + .... + ( x+99 ) = 0

           =x.50 + ( 1+3+ ... +99 ) = 0

=) SSH: (99-1):2+1=50

   SC: 50:2=25

   TMC: 99+1=100

   T: 100.25 = 2500

=) x.50 + 2500 = 0 

   x.50 = 0-2500 = -2500

  x         = -2500:50 = - 50

vậy x=-50

Câu 1: D

Câu 2: C

Câu 3: B

325253737747⁸⁹⁰⁷⁶⁵⁴³ chuyển đổi sang STN là?

1, để \(\dfrac{2x+1}{x+3}\) là 1 số nguyên 

= > 2x + 1 chia hết cho x + 3 ( x thuộc Z và x \(\ne3\) )

= > 2 ( x + 3 ) - 5 chia hết cho x + 3 

=> -5 chia hết cho x + 3 

hay x + 3 thuộc Ư(-5 ) \(\in\left\{\pm1;\pm5\right\}\)

Đến đây em tự tìm các giá trị của x

2, Tương tự câu 1, x - 1 chia hết cho x + 5 ( x thuộc Z và x khác - 5 )

= > - 6 chia hết cho x + 5 

= > \(x+5\in\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

....

3,  ( x - 1 ) ( y - 3 ) = 7 

x,y thuộc Z = > x - 1 ; y - 3 thuộc Ư(7)

và ( x - 1 )( y - 3 ) = 7

( 1 ) \(\left\{{}\begin{matrix}x-1=1\\y-3=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=10\end{matrix}\right.\)

(2) \(\left\{{}\begin{matrix}x-1=7\\y-3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=4\end{matrix}\right.\)

( 3) \(\left\{{}\begin{matrix}x-1=-1\\y-3=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-4\end{matrix}\right.\)

( 4 ) \(\left\{{}\begin{matrix}x-1=-7\\y-3=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=2\end{matrix}\right.\)

Từ ( 1 ) , ( 2 ) , ( 3 ) , ( 4 ) các cặp giá trị ( x,y ) nguyên cần tìm là ....

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

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

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

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

`<=>(x-3)(x+3)=0`

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

`<=>(x+5)(x-1)=0`

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

1.a.

\(\left(x+1\right)\left(x+2\right)\left(x-2\right)\left(x+5\right)\ge m\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x^2+3x-10\right)\ge m\)

Đặt \(x^2+3x-10=t\ge-\dfrac{49}{4}\)

\(\Rightarrow\left(t+2\right)t\ge m\Leftrightarrow t^2+2t\ge m\)

Xét \(f\left(t\right)=t^2+2t\) với \(t\ge-\dfrac{49}{4}\)

\(-\dfrac{b}{2a}=-1\) ; \(f\left(-1\right)=-1\) ; \(f\left(-\dfrac{49}{4}\right)=\dfrac{2009}{16}\)

\(\Rightarrow f\left(t\right)\ge-1\)

\(\Rightarrow\) BPT đúng với mọi x khi \(m\le-1\)

Có 30 giá trị nguyên của m

1b.

Với \(x=0\)  BPT luôn đúng

Với \(x\ne0\) BPT tương đương:

\(\dfrac{\left(x^2-2x+4\right)\left(x^2+3x+4\right)}{x^2}\ge m\)

\(\Leftrightarrow\left(x+\dfrac{4}{x}-2\right)\left(x+\dfrac{4}{x}+3\right)\ge m\)

Đặt \(x+\dfrac{4}{x}-2=t\) \(\Rightarrow\left[{}\begin{matrix}t\ge2\\t\le-6\end{matrix}\right.\)

\(\Rightarrow t\left(t+5\right)\ge m\Leftrightarrow t^2+5t\ge m\)

Xét hàm \(f\left(t\right)=t^2+5t\) trên \(D=(-\infty;-6]\cup[2;+\infty)\)

\(-\dfrac{b}{2a}=-\dfrac{5}{2}\notin D\) ; \(f\left(-6\right)=6\) ; \(f\left(2\right)=14\)

\(\Rightarrow f\left(t\right)\ge6\)

\(\Rightarrow m\le6\)

Vậy có 37 giá trị nguyên của m thỏa mãn

2.

Xét với \(x\ge1\)

\(m\left(x+1\right)+3\left(x-1\right)-2\sqrt{x^2-1}=0\)

\(\Leftrightarrow m+3\left(\dfrac{x-1}{x+1}\right)-2\sqrt{\dfrac{x-1}{x+1}}=0\)

Đặt \(\sqrt{\dfrac{x-1}{x+1}}=t\Rightarrow0\le t< 1\)

\(\Rightarrow m+3t^2-2t=0\)

\(\Leftrightarrow3t^2-2t=-m\)

Xét hàm \(f\left(t\right)=3t^2-2t\) trên \(D=[0;1)\)

\(-\dfrac{b}{2a}=\dfrac{1}{3}\in D\) ; \(f\left(0\right)=0\) ; \(f\left(\dfrac{1}{3}\right)=-\dfrac{1}{3}\) ; \(f\left(1\right)=1\)

\(\Rightarrow-\dfrac{1}{3}\le f\left(t\right)< 1\)

\(\Rightarrow\) Pt có nghiệm khi \(-\dfrac{1}{3}\le-m< 1\)

\(\Leftrightarrow-1< m\le\dfrac{1}{3}\)

a)\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{8}{12}\)

\(\dfrac{2}{5}+x=\dfrac{3}{12}\)

\(\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(x=\dfrac{5}{20}-\dfrac{8}{20}\)

\(x=\dfrac{-3}{20}\)

b)\(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow2x=0\) hoặc \(x-\dfrac{1}{7}=0\)

\(x=0:2\) \(x=0+\dfrac{1}{7}\)

\(x=0\) \(x=\dfrac{1}{7}\)

\(\Rightarrow x=0\) hoặc \(x=\dfrac{1}{7}\)

c)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\dfrac{1}{4}:x=\dfrac{8}{20}-\dfrac{15}{20}\)

\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)

\(x=\dfrac{1}{4}.\dfrac{-20}{7}\)

x= \(\dfrac{1.\left(-5\right)}{1.7}\)

\(x=\dfrac{-5}{7}\)

Mệnh đề A sai, sửa :

\(x\in\left[1;3\right]\Leftrightarrow1\le x\le3\)

Mệnh đề C sai, sửa :

\(x\in\left(-\infty;3\right)\)

Mệnh đề D sai, sửa :

\(x\in[1;3)\Leftrightarrow1\le x< 3\)

\(\Rightarrow\)Mệnh đề B đúng

\(.\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

\(\Rightarrow x-7=0\)

\(\Rightarrow x=7\)

Vậy : x=7