a: =>1/3x-2/5x=5

=>-1/15x=5

=>x=-75
b: =>4x=4

=>x=1

c: =>6*3^x-5*3^x=243

=>3^x=243

=>x=5

\(a,\Rightarrow5x+3x^2-3x^2-x+2=6\\ \Rightarrow4x=4\Rightarrow x=1\\ b,\Rightarrow\left(2x+\dfrac{1}{2}-1+2x\right)\left(2x+\dfrac{1}{2}+1-2x\right)=2\\ \Rightarrow\dfrac{3}{2}\left(4x-\dfrac{1}{2}\right)=2\\ \Rightarrow6x-\dfrac{3}{4}=2\\ \Rightarrow6x=\dfrac{11}{4}\\ \Rightarrow x=\dfrac{11}{24}\\ c,\Rightarrow\left(x+3\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

Bài 1:

\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)

Bài 2:

\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)

a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)

\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)

\(\Leftrightarrow-12x=12\)

hay x=-1

a) \(2^x\cdot4=128\)

\(\Rightarrow2^x\cdot2^2=2^7\)

\(\Rightarrow2^{x+2}=2^7\)

\(\Rightarrow x+2=7\)

\(\Rightarrow x=5\)

b) \(\left(2x+1\right)^3=125\)

\(\Rightarrow\left(2x+1\right)^3=5^3\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=4:2\)

\(\Rightarrow x=2\)

c) \(2x-2^6=6\)

\(\Rightarrow2x-64=6\)

\(\Rightarrow2x=70\)

\(\Rightarrow x=70:2\)

\(\Rightarrow x=35\)

d) \(64\cdot4^x=45\)

\(\Rightarrow4^3\cdot4^x=45\)

\(\Rightarrow4^{x+3}=45\)

Xem lại đề

e) \(27\cdot3^x=243\)

\(\Rightarrow3^3\cdot3^x=3^5\)

\(\Rightarrow3^{x+3}=3^5\)

\(\Rightarrow x+3=5\)

\(\Rightarrow x=2\)

g) \(49\cdot7^x=2401\)

\(\Rightarrow7^2\cdot7^x=7^4\)

\(\Rightarrow7^{x+2}=7^4\)

\(\Rightarrow x+2=4\)

\(\Rightarrow x=2\)

h) \(3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

k) \(3^4\cdot3^x=3^7\)

\(\Rightarrow3^{x+4}=3^7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

n) \(3^x+25=26\cdot2^2+2\cdot3^0\)

\(\Rightarrow3^x+25=104+2\)

\(\Rightarrow3^x+25=106\)

\(\Rightarrow3^x=81\)

\(\Rightarrow3^x=3^4\)

\(x=4\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`2^x*4 = 128`

`=> 2^x = 128 \div 4`

`=> 2^x = 2^7 \div 2^2`

`=> 2^x = 2^5`

`=> x = 5`

Vậy, `x = 5.`

`b)`

\(\left(2x+1\right)^3=125\)

`=> (2x + 1)^3 = 5^3`

`=> 2x + 1 = 5`

`=> 2x = 5-1`

`=> 2x = 4`

`=> x = 4 \div 2`

`=> x = 2`

Vậy, `x = 2`

`c)`

\(2x-2^6=6\)

`=> 2x = 6+2^6`

`=> 2x = 70`

`=> x = 70 \div 2`

`=> x = 35`

Vậy, `x = 35`

`d)`

\(64\cdot4^x=45\) Bạn xem lại đề

`e)`

`27*3^x = 243`

`=> 3^3 * 3^x = 3^5`

`=> 3^(3 + x) = 3^5`

`=> 3 + x = 5`

`=> x = 5 - 3`

`=> x = 2`

Vậy, `x = 2`

`g)`

`49* 7^x = 2401`

`=> 7^2 * 7^x = 7^4`

`=> 7^(2 + x) = 7^4`

`=> 2 + x = 4`

`=> x = 4 - 2`

`=> x = 2`

Vậy, `x = 2`

`h)`

`3^x = 81`

`=> 3^x = 3^4`

`=> x = 4`

Vậy, `x = 4`

`k)`

`3^4 * 3^x = 3^7`

`=> 3^(4 + x) = 3^7`

`=> 4 + x = 7`

`=> x = 7 - 4`

`=> x = 3`

Vậy, `x = 3`

`n)`

`3^x + 25 = 26*2^2 + 2*3^0`

`=> 3^x + 25 = 104 + 2`

`=> 3^x + 25 = 106`

`=> 3^x = 106 - 25`

`=> 3^x = 81`

`=> 3^x = 3^4`

`=> x = 4`

Vậy, `x = 4.`

\(#48Cd\)

a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)

\(\Rightarrow3x-6x^2+6x+14=29\)

\(\Rightarrow-6x^2+9x-15=0\)

\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)

\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)

Vậy \(S=\varnothing\)

a. \(2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

hay x=-2

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

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

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

a) \(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)

b) \(\Rightarrow\left(2x+1\right)^3=5^3\)

\(\Rightarrow2x+1=5\Rightarrow x=2\)

c) \(\Rightarrow2^x=32\Rightarrow x=5\)

d) \(\Rightarrow4^3.4^x=4^5\Rightarrow4^x=4^2\Rightarrow x=2\)

e) \(\Rightarrow3^3.3^x=3^5\Rightarrow3^x=3^2\Rightarrow x=2\)

f) \(\Rightarrow7^2.7^x=7^4\Rightarrow7^x=7^2\Rightarrow x=2\)

a. 2^x . 4 = 128

<=> 2^{x + 2} = 2⁷

<=> x + 2 = 7

<=> x = 5

b. (2x + 1)³ = 125

<=> (2x + 1)³ - 5³ = 0

<=> (2x + 1 - 5)\(\left[\left(2x+1\right)^2+\left(2x+1\right).5+25\right]=0\)

<=> (2x - 4)(4x² + 4x + 1 + 10x + 5 + 25) = 0

<=> (2x - 4)(4x² + 14x + 31) = 0

<=> \(\left[{}\begin{matrix}2x-4=0\\4x^2+14x+31=0\end{matrix}\right.\)

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

c. 2^x - 26 = 6

<=> 2^x = 32

<=> x = 5

d. 64 . 4^x = 4⁵

<=> 4³ . 4^x = 4⁵

<=> 4^{3 + x} = 4⁵

<=> 3 + x = 5

<=> x = 2

e. 27 . 3^x = 243

<=> 3³ . 3^x = 3⁵

<=> 3^{3 + x} = 3⁵

<=> 3 + x = 5

<=> x = 2

g. 49 . 7^x = 2401 (Bn xem lại đề câu này)

<=> 7² . 7^x = 7⁴

<=> 7^{2 + x} = 7⁴

<=> 2 + x = 4

<=> x = 2

a, 2^x.4= 128

=> 2^x= 128:4

=>2^x= 32

=> 2^x=2⁵

=> x=5

\(\left(2x+1\right)^3=125\)

\(\Rightarrow\left(2x+1\right)^3=\left(\pm5\right)^3\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c, 2^x - 26 = 6

=> 2^x = 32

=> 2^x= 2⁵

=> x=5 

d, 64. 4^x = 4⁵

=> 4³. 4^x= 4⁵

=> 3+x=5

=> x=2

e, 27. 3^x= 243

=> 3³. 3^x= 3⁵

=> 3+x=5

=> x=2

Hoctot