Bài 5:

a)x+37=50

⇔x=13

b)2x-3=11

⇔2x=14

⇔x=7

c)(2+x):5=6

⇔2+x=30

⇔x=28

d)2+x:5=6

⇔x:5=4

⇔x=20

a)x+37=50
   x      =50-37
  x       =13
b)2.x-3=11
   2.x   =11+3
   2.x   =14
      x   =14:2
      x   =7
c)(2+x):5=6
   2+x =6.5
   2+x =30
  x=30-2
  x=28
d)2+x:5=6
   x:5=6-2
   x:5=4
   x=4.5
   x=20
Câu trả lời nhớ ghi đầy đủ như này nha bạn

   

\(a,x< 50\Leftrightarrow\sqrt{x}-1< 5\sqrt{2}-1\\ M=\dfrac{\sqrt{x}-1}{2}\in Z\\ \Leftrightarrow\sqrt{x}-1\in B\left(2\right)=\left\{0;2;4;6\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{1;3;5;7\right\}\\ \Leftrightarrow x\in\left\{1;9;25;49\right\}\\ b,\Leftrightarrow\sqrt{x}-5\inƯ\left(9\right)=\left\{-3;-1;1;3;9\right\}\left(\sqrt{x}-5>-5\right)\\ \Leftrightarrow\sqrt{x}\in\left\{2;4;6;8;14\right\}\\ \Leftrightarrow x\in\left\{4;16;36;64;196\right\}\)

b: 30 chia hết cho x

45 chia hết cho x

Do đó: \(x\inƯC\left(30;45\right)=Ư\left(15\right)\)

mà x>10

nen x=15

c: \(\Leftrightarrow x+2\in\left\{1;-1;3;-3;9;-9\right\}\)

hay \(x\in\left\{-1;-3;1;-5;7;-11\right\}\)

d: =>x+3+14 chia hết cho x+3

=>\(x+3\in\left\{1;-1;2;-2;7;-7;14;-14\right\}\)

hay \(x\in\left\{-2;-4;-1;-5;4;-10;11;-17\right\}\)

\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)

\(a,\left(x-4\right).5=50\\ \Rightarrow x-4=10\\ \Rightarrow x=14\\ b,\left(-470-x\right)\left(-3\right)=-120\\ \Rightarrow-470-x=40\\ \Rightarrow x=-510\\ c,5x^3+\left(-170\right)=-210\\ \Rightarrow5x^3=-40\\ \Rightarrow x^3=-8\\ \Rightarrow x=-2\)

a,(x−4).5=50⇒x−4=10⇒x=14

b,(−470−x)(−3)=−120⇒−470−x=40⇒x=−510

c,5x3+(−170)=−210⇒5x3=−40⇒x3=−8⇒x=−2

x - 4 = 50. 5

x - 4 = 250

x = 250 + 4

x = 254

Vậy x = 254

b) -470 - x = -120. (-3)

-470 - x = 360

x = -470 - 360 

x = -830

Vậy x = -830

c) 5. x³ = -210 - (-170)

5. x³ = -40

x³ = -40: 5

x³ = -8

x = -2

Vậy x = -2

a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)

\(\Leftrightarrow-62x=-92\)

hay \(x=\dfrac{46}{31}\)

b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)

\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)

\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)

\(\Leftrightarrow8x^2+49x-15=0\)

\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là: 

\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)

a: x chia hết cho 4;5;10

nên \(x\in BC\left(4;5;10\right)\)

mà 10<=x<50

nên x=40

b: x=33

a: =>(x-2)^3*[(x-2)^8-1]=0

=>(x-2)(x-3)(x-1)=0

=>\(x\in\left\{2;3;1\right\}\)

b: (x-5)^24=(x-5)^9

=>\(\left(x-5\right)^9\cdot\left[\left(x-5\right)^{15}-1\right]=0\)

=>x-5=0 hoặc x-5=1

=>x=6 hoặc x=5

c: =>(x-5)^4*[(x-5)^21-1]=0

=>x-5=0 hoặc x-5=1

=>x=5 hoặc x=6

a) \(\left(x-2\right)^{11}=\left(x-2\right)^3\)

\(\Rightarrow\left(x-2\right)^{11}-\left(x-2\right)^3=0\)

\(\Rightarrow\left(x-2\right)^3\left[\left(x-2\right)^8-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^3=0\\\left(x-2\right)^8-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x-2\right)^8=1\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

b) \(\left(x-5\right)^{24}=\left(x-5\right)^9\)

\(\Rightarrow\left(x-5\right)^{24}-\left(x-5\right)^9=0\)

\(\Rightarrow\left(x-5\right)^9\left[\left(x-5\right)^{15}-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^9=0\\\left(x-5\right)^{15}-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{15}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

c) \(\left(x-5\right)^{25}=\left(x-5\right)^4\)

\(\Rightarrow\left(x-5\right)^{25}-\left(x-5\right)^4\)

\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^{21}-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\\left(x-5\right)^{21}-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{21}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

a, Ta có : 

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

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

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)

\(\Rightarrow x=11;y=17;z=23\)

b, Đặt \(\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\Rightarrow xyz=810\)

\(\Rightarrow2k.3k.5k=810\Leftrightarrow30k^3=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)

\(\Rightarrow x=6;y=9;z=15\)

a) Ta có: \(\dfrac{x-1}{2}=\dfrac{2x-2}{4};\dfrac{y-2}{3}=\dfrac{3y-6}{9};\dfrac{z-3}{4}\)

Áp dụng t/c dtsbn:

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=12\end{matrix}\right.\)

b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

xyz = 810

=> 2k.3k.5k = 810

=> k = 3

\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)

a) Ta có: \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)

nên \(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

mà 2x+3y-z=50

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)

Do đó:

\(\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)

b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Ta có: xyz=810

\(\Leftrightarrow30k^3=810\)

\(\Leftrightarrow k^3=27\)

\(\Leftrightarrow k=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\cdot3=6\\y=3k=3\cdot3=6\\z=5k=5\cdot3=15\end{matrix}\right.\)

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