a) 

B(14) = 0; 14; 28; 42; 56; 70; 84; …..

Vì 20 < x < 80 => x ∈ { 28; 42; 56; 70．}

b)

Vì 70 chia hết cho x ѵà 80 chia hết cho x => x ∈ ƯC(70; 80)

Phân tích:

70 = 2 ．5 ．7

80 = 24 ．5

ƯCLN (70; 80) = 2.5=10

ƯC ( 70; 80) = Ư(10) ={1;2;5;10}

Mà x > 8 => x = 10

c)

Vì 126 chia hết cho x ѵà 210 chia hết cho x => x ∈ ƯC(126; 210)

Phân tích

126 = 2 ．3² ．7

210 = 2 ．3 ．5 ．7

ƯCLN(126; 210) = 2 ．3 ．7 = 42

ƯC(126; 210) = { 1; 2; 3; 6; 7; 14; 21; 42 }

Vì 15 < x < 30 => x = 21

a) \(B\left(14\right)=\left\{0;14;28;42;56;70;84;.......\right\}\)

Vì \(20< x< 8\Rightarrow x\in\left\{28;42;56;70\right\}\)

b) Vì 70 chia hết cho x và 80 chia hết cho x nên \(\Rightarrow x\inƯC\left(70;80\right)\)

Phân tích :

\(70=2.5.7\)

\(80=2^4.5\)

Mà \(x>8\Rightarrow x=10\)

c) Vì 126 chia hết cho x  và 210 chia hết cho x nên \(\Rightarrow x\inƯC\left(126;210\right)\)

Phân tích :

\(126=2.3^2.7\)

\(210=2.3.5.7\)

\(ƯC\left(126;210\right)=\left\{1;2;3;6;7;14;21;42\right\}\)

\(ƯCLN\left(126;210\right)=2.3.7=42\)

Theo đề : \(x\in\left(>15< 30\right)\Rightarrow x=21\)

\(a,\Rightarrow6x+70=130\Rightarrow6x=60\Rightarrow x=10\\ b,\Rightarrow240=\left(x+70\right):14-40\\ \Rightarrow\left(x+70\right):14=280\\ \Rightarrow x+70=3920\Rightarrow x=3850\\ c,\Rightarrow x-15=75\Rightarrow x=90\\ d,\Rightarrow\left(x+175\right):5=680-30=650\\ \Rightarrow x+175=3250\Rightarrow x=3075\\ e,\Rightarrow x-4867=1004523\Rightarrow x=1009390\\ f,\Rightarrow x+32-17=24\Rightarrow x=9\\ g,\Rightarrow3\left(x+1\right)=54\Rightarrow x+1=18\Rightarrow x=17\\ h,\Rightarrow19\left(35:x+3\right)=152\\ \Rightarrow35:x+3=8\Rightarrow35:x=11\Rightarrow x=\dfrac{35}{11}\)

a) \(\sqrt{\left(2x-3\right)^2}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)

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

b) \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\left(đk:x\ge-2\right)\)

\(\Leftrightarrow8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}=20\)

\(\Leftrightarrow5\sqrt{x+2}=20\)

\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)

c) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)

\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

a. \(\sqrt{\left(2x-3\right)^2}=7\)

<=> \(\left|2x-3\right|=7\)

<=> \(\left[{}\begin{matrix}2x-3=7\left(x\ge\dfrac{3}{2}\right)\\-2x+3=7\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}2x=10\\-2x=4\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=5\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)

b. \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\)  ĐK: \(x\ge-2\)

<=> \(\sqrt{64\left(x+2\right)}-\sqrt{25\left(x+2\right)}+\sqrt{4\left(x+2\right)}-20=0\)

<=> \(8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}-20=0\)

<=> \(\sqrt{x+2}.\left(8-5+2\right)-20=0\)

<=> \(5\sqrt{x+2}=20\)

<=> \(\sqrt{x+2}=4\)

<=> \(\left(\sqrt{x+2}\right)^2=4^2\)

<=> \(\left|x+2\right|=16\)

<=> \(\left[{}\begin{matrix}x+2=16\left(x\ge-2\right)\\x+2=-16\left(x< -2\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=14\left(TM\right)\\x=-18\left(TM\right)\end{matrix}\right.\)

c. \(\sqrt{x^2-9}-3\sqrt{x-3}=0\)             ĐK: \(x\ge3\)

<=> \(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)

<=> \(\sqrt{x-3}.\sqrt{x+3}-3\sqrt{x-3}=0\)

<=> \(\left(\sqrt{x+3}-3\right).\sqrt{x-3}=0\)

<=> \(\left[{}\begin{matrix}\sqrt{x+3}-3=0\\\sqrt{x-3}=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=6\\x=3\end{matrix}\right.\)

a) I2x-3I=7

        2x-3=7     =>x=5

        2x-3=-7     =>x=-2

b) \(8\sqrt{3x}-5\sqrt{3x}+2\sqrt{3x}=20\)

   5\(\sqrt{3x}=20\)

   3x=16  =>x=16/3

c)  vì câu c dài nên mình chỉ cho đáp án thôi là 0,3,6

vì \(\sqrt{ }\) của 1 số luôn dương nên 3,6 thỏa mãn 

Bài 10:

a) (x+2)² -x(x+3) + 5x = -20

=> x² + 4x + 4 - x² - 3x + 5x = -20

=> 6x = -20 + (-4)

=> 6x = -24

=> x = -4

b) 5x³-10x²+5x=0   

=>5x(x²-2x+1)=0

=>5x(x-1)² =0

=> 5x=0 hoặc (x-1)²=0

=>x=0 hoặc x=1

c) (x² - 1)³ - (x⁴ + x² + 1)(x² - 1) = 0

=> (x² - 1)[(x² - 1)² -  (x⁴ + x² + 1)] = 0

<=> (x² - 1)(x⁴ - 2x² + 1 - x⁴ - x² - 1) = 0

<=>  (x² - 1)(-3x²) = 0

<=> (x² - 1)=0 hoặc (-3x²) =0

<=> x²=1 hoặc x²=0

<=> x=−1;1 hoặc x=0

d)

⇔x=-23/12

Học tốt nhé:333

\(x^2+3x=0\)

\(\Leftrightarrow x\left(x+3\right)=0\)

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

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

\(x^2+3x=0\)

\(x\left(x+3\right)=0\)

x = 0 hoặc x +3 = 0

=> x = 0 hoặc x = -3

Vậy ...

x(x+3)=0

=> x=0 hoặc x=-3

=> x thuộc { 0; 3}

a) \(0,\left(31\right)+x=0,3\left(7\right)\\ \Rightarrow\dfrac{31}{99}+x=\dfrac{17}{45}\\ \Rightarrow x=\dfrac{17}{45}-\dfrac{31}{99}=\dfrac{32}{495}=0,0\left(64\right)\)

Vậy \(x=0,0\left(64\right)\)

b) \(0,\left(4\right)\cdot x=\dfrac{5}{6}\\ \Rightarrow\dfrac{4}{9}\cdot x=\dfrac{5}{6}\\ \Rightarrow x=\dfrac{5}{6}:\dfrac{4}{9}\\ \Rightarrow x=\dfrac{5}{6}\cdot\dfrac{9}{4}\\ \Rightarrow x=\dfrac{15}{8}=1,875\)

Vậy \(x=1,875\)

a) x= + - 5
b) x\(\in\)\(\left\{-1;-7\right\}\)

a/ \(x^2-25=0\)

\(\Rightarrow\left(x+5\right)\left(x-5\right)=0\)

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

b/ \(x\left(x+7\right)+x+7=0\)

\(x\left(x+7\right)+\left(x+7\right)=0\)

\(\left(x+7\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+7=0\Rightarrow x=-7\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)

a) x^2 - 25 = 0 <=> (x + 5)(x - 5) = 0

<=> x = -5 hoặc x = 5

b) x(x + 7) + x + 7 = 0 <=> (x + 7)(x + 1) = 0

<=> x = -7 hoặc x = -1

b) \(x^3-x^2-x+1=0\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)

\(\Leftrightarrow x-1=0\)   hoặc   \(x+1=0\)

\(\Leftrightarrow x=1\)         hoặc    \(x=-1\)

c) \(x^2-6x+8=0\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

a) \(x^3+x^2+x+1=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

(do \(x^2+1\ge1>0\))

a: Ta có: \(x^3+x^2+x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

b: Ta có: \(x^3-x^2-x+1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

c: Ta có: \(x^2-6x+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

a) 2436 : x = 12

\(\Rightarrow\) x = 2436 : 12

\(\Rightarrow\) x = 203

b) 6x - 5 = 613

\(\Rightarrow\) 6x = 618

\(\Rightarrow\) x = 103

c) 12(x - 1) = 0

\(\Rightarrow\) x - 1 = 0

\(\Rightarrow\) x = 1

d) 0 : x = 0    (đúng \(\forall\)x  \(\in\) N)

\(\Rightarrow\) x \(\in\) N

a) Ta có: 2436:x=12

nên x=2436:12

hay x=203

b) Ta có: 6x-5=613

nên 6x=618

hay x=103

c) Ta có: 12(x-1)=0

mà 12>0

nên x-1=0

hay x=1

d) Ta có: 0:x=0

nên \(x\in R;x\ne0\)

a.2436:x=12

            x=2436:12

            x=203

b.6x-5=613

   6x   =613+5

   6x   =618

      x  =618:6

       x=103

c, 12(x-1)=0

          x-1=0:12

           x-1=0

              x= 0+1

              x=1

d, x:0=0

Vì theo quy luật, 0 chia số nào cx bằng 0. Suy ra x là số bất kỳ.

Nhớ like cho mik nha