nik lộn

2^x+1. 2^{2009 =} 2²⁰¹⁰

10 - 2x = 25 - 3x

a: =-3/4-1/4+2/7+5/7+2023/2024

=-1+1+2023/2024=2023/2024

b: 2/3x=2/7

=>x=2/7:2/3=3/7

c; =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5

=>x=3/5:2/3=3/5*3/2=9/10

a: \(\dfrac{x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)

\(=\dfrac{x+10}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-\left(x+10\right)}{2\left(x+2\right)}\)

b: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)

\(=\dfrac{\left(2x-1\right)\left(2x+1\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(x-2\right)}\)

\(=\dfrac{3\left(2x-1\right)\left(2x+1\right)}{2\left(x-2\right)\left(x+4\right)}\)

c: \(=\dfrac{4y^2}{7x^4}\cdot\dfrac{35x^2}{-8y}=\dfrac{5}{x^2}\cdot\dfrac{-1}{2}\cdot y=\dfrac{-5y}{2x^2}\)

d: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)

a) (x+3)² + (4+x)(4-x) = 10

x² + 6x + 9 + 16- x² = 10

6x + 25 = 10

6x = -15

x = -15/6

b) 9(x+1)² - (3x-2)(3x+2) = 10

9x² + 18x + 9 - 9x² + 4 =10

18x + 13 = 10

18x = -3

x = -1/6

a) ( x + 3 )² + ( 4 + x )( 4 - x ) = 10

⇔ x² + 6x + 9 + 16 - x² = 10

⇔ 6x + 25 = 10

⇔ 6x = -15

⇔ x = -15/6 = -5/2

b) 9( x + 1 )² - ( 3x - 2 )( 3x + 2 ) = 10

⇔ 9( x² + 2x + 1 ) - ( 9x² - 4 ) = 10

⇔ 9x² + 18x + 9 - 9x² + 4 = 10

⇔ 18x + 13 = 10

⇔ 18x = -3

⇔ x = -3/18 = -1/6

a) \(\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=10\)

\(x^2+6x+9+16-x^2=10\)

\(6x+25=10\)

\(6x=-15\)

\(x=-\frac{5}{2}\)

b) \(9.\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)

\(9.\left(x^2+2x+1\right)-9x^2+4=10\)

\(9x^2+18x+9-9x^2+4=10\)

\(18x+13=10\)

\(18x=-3\)

\(x=-\frac{1}{6}\)

c)

\(4\left(3x-4\right)-2=18\)

<=> \(12x-16-2=18\)
<=> \(12x=36\)

<=> \(x=3\)

Vậy x=3

d)

\(\left(3x-10\right):10=50\)

<=> \(3x-10=500\)

<=> \(3x=510\)

<=> x= \(170\)

Vậy x= 170

f)

\(x-\left[42+\left(-25\right)\right]=-8\)

<=> \(x-17=-8\)

<=> x= \(9\)

Vậy x=9

h)

\(x+5=20-\left(12-7\right)\)

<=> \(x+5=15\)

<=> \(x=10\)

Vậy x= 10

k)

\(\left|x-5\right|=7-\left(-3\right)\)

<=> \(\left|x-5\right|=10\)

* Với \(x>=5\) ; ta được:

\(x-5=10\)

<=> x= 15 (thoả mãn điều kiện )

*Với \(x< 5\) ; ta được:

\(-\left(x-5\right)=10\)

<=> \(-x+5=10\)

<=> \(-x=5\)

<=> \(x=-5\) (thoả mãn điều kiện)

Vậy x=15 ; x= -5

i)

\(\left|x-5\right|=\left|7\right|\)

<=> \(\left|x-5\right|=7\)

*Với \(x>=5\) ; ta được:

\(x-5=7\)

<=> \(x=12\) (thoả mãn)

*Với \(x< 5\) ; ta được:

\(-\left(x-5\right)=7\)

<=> \(-x=2\)

<=> \(x=-2\) (thoả mãn)

Vậy x= 12; x= -2

m)

\(2^{x+1}.2^{2009}=2^{2010}\)

<=> \(2^{x+1+2009}=2^{2010}\)

<=> \(2^{x+2010}=2^{2010}\)

=> \(x+2010=2010\)

=> \(x=0\)

Vậy x=0

n)

\(10-2x=25-3x\)

<=>\(x=15\)

Vậy x=15

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

6) ĐKXĐ: \(x\le-6\)

\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)

\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)

Vậy \(x\le-6\)

7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)

\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)

\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)

Vậy \(x\ge\dfrac{2}{3}\)

8) ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)

\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)

9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

10-2(4-3x) = -4

=> 2(4-3x) = 10-(-4)

=> 2(4-3x) = 14

=> 4-3x = 14:2

=> 4-3x = 7

=> 3x = 4-7

=> 3x = -3

=> x = -1

Vậy x = -1

Chúc bạn học tốt !

\(10-2\left(4-3x\right)=-4\)

\(10-2\left(4-3x\right)=2\left(3x+1\right)\)

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

\(6x+2=-4\)

\(6x=-6\)

\(\Rightarrow x=-1\)