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b: Ta có: \(\left(2x+7\right)^2=9\left(x+2\right)^2\)

\(\Leftrightarrow\left(3x+4-2x-7\right)\left(3x+4+2x+7\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{5}\end{matrix}\right.\)

c: ta có: \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)

\(\Leftrightarrow\left(3x-6\right)^2-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(3x-6-x-2\right)\left(3x-6+x+2\right)=0\)

\(\Leftrightarrow\left(2x-8\right)\left(4x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)

13 tháng 12 2022

a.1+3+5+7+...+2x+1=225

[(2x+1-1):2+1]x(2x+1+1):2=225

(x+1)x(2x+2):2=225

(x+1)x(x+1)=225

(x+1)2=225

(x+1)2=152

x+1=15

x=14

_______________

b. 130-[5.(9-x)+43]=47

    5.(9-x)+43=83

  5.(9-x)=40

9-x=8

x=1

_______________

c.16x<324

24x<220

=>x∈{0;1;2;3;4}

13 tháng 12 2022

a: =>(x+1)^2=225

=>x+1=15

=>x=14

b: =>[5*(9-x)+43]=130-47=83

=>5(9-x)=40

=>9-x=8

=>x=1

c: =>2^4x<2^20

=>4x<20

=>0<x<5

13 tháng 7 2021

a, ( 2x - 3 )2- (2x + 1)2 = -3

4x2-12x+9-4x2+4x-1=-3

-8x-1=-3

-8x=-2

x=\(\frac{1}{4}\)

b, (5x - 1) 2 - (5x + 4)(5x - 4) = 7

25x2-10x+1-25x2+16=7

-10x+17=7

-10x=-10

x=1

c, ( x- 5)2 + (x-3)(x+3) - 2(x + 1)2=0

x2-10x+25+x2-9-2x2-4x-2=0

-14x+14=0

-14(x-1)=0

=>x-1=0

x=1

13 tháng 7 2021

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

\(\Leftrightarrow4x^2-12x+9-4x^2-4x-1=-3\)

\(\Leftrightarrow-16x+8=-3\)

\(\Leftrightarrow-16x=-11\)

\(\Leftrightarrow x=\frac{11}{16}\)

b)\(\left(5x-1\right)^2-\left(5x+4\right)\left(5x-4\right)=7\)

\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)

\(\Leftrightarrow-10x+17=7\)

\(\Leftrightarrow-10x=-10\)

\(\Leftrightarrow x=1\)

c)\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)

\(\Leftrightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow2x^2-10x-16-2x^2-4x-2=0\)

\(\Leftrightarrow-14x-18=0\)

\(\Leftrightarrow-14x=18\)

\(\Leftrightarrow x=-\frac{9}{7}\)

#H

17 tháng 3 2023

ỏ cảm mơn nhaaaa ! có j giúp típ nha thank kiuuu 

5 tháng 10 2020

a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)

\(\Leftrightarrow2x=-40\)

\(\Rightarrow x=-20\)

b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow x^3+27-x^3+4x=15\)

\(\Leftrightarrow4x=-12\)

\(\Rightarrow x=-3\)

c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)

\(\Leftrightarrow-14x=14\)

\(\Rightarrow x=-1\)

5 tháng 10 2020

d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)

\(\Leftrightarrow17x=-34\)

\(\Rightarrow x=-2\)

e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Leftrightarrow24x=24\)

\(\Rightarrow x=1\)