mk chi lam duoc cau c thoi

suy ra 2x-1=5 hoac 2x-1=-5

          2x=6            2x=-4

          x=3                 x=-2

vay x=3 hoac x=-2

a. 3/4.x+1/3=-1/2

=>3/4.x=-1/2-1/3=-5/6

=>x=-5/6 chia 3/4=-10/9

b. -x/4=-9/x =>-x*x=4*-9

=>-2x=-36 =>x=18

c./2x-1/=5

=> 2x-1=5 =>2x=5+1=6 =>x=3

  hoặc 2x-1=-5 =>2x=-5+1=-4 =>x=-2

d,e: Sai đề rồi

a)

Đặt \(\frac{x}{2}=t\Rightarrow 3^{2t}-4=5^t\)

\(\Leftrightarrow 9^t-5^t=4\)

TH1: \(t>1\Rightarrow 9^t-5^t< 4^t\)

\(\Leftrightarrow 9^t< 4^t+5^t\)

\(\Leftrightarrow 1< \left(\frac{4}{9}\right)^t+\left(\frac{5}{9}\right)^t\) \((*)\)

Ta thấy vì \(\frac{4}{9};\frac{5}{9}<1 \), do đó với \(t>1\Rightarrow \left\{\begin{matrix} \left(\frac{4}{9}\right)^t< \frac{4}{9}\\ \left(\frac{5}{9}\right)^t< \frac{5}{9}\end{matrix}\right.\)

\(\Rightarrow \left(\frac{4}{9}\right)^t+\left(\frac{5}{9}\right)^t< \frac{4}{9}+\frac{5}{9}=1\) (mâu thuẫn với (*))

TH2: \(t<1 \) tương tự TH1 ta cũng suy ra mâu thuẫn

do đó \(t=1\Rightarrow x=2\)

b)

Ta có: \(5^{2x}=3^{2x}+2.5^x+2.3^x\)

\(\Leftrightarrow (5^{2x}-2.5^{x}+1)=3^{2x}+2.3^x+1\)

\(\Leftrightarrow (5^x-1)^2=(3^x+1)^2\)

\(\Leftrightarrow (5^x-3^x-2)(5^x+3^x)=0\)

Dễ thấy \(5^x+3^x>0\forall x\in\mathbb{R}\Rightarrow 5^x-3^x-2=0\)

\(\Leftrightarrow 5^x-3^x=2\)

\(\Leftrightarrow 5^x=3^x+2\)

Đến đây ta đưa về dạng giống hệt phần a, ta thu được nghiệm \(x=1\)

c)

\((2-\sqrt{3})^x+(2+\sqrt{3})^x=4^x\)

\(\Leftrightarrow \left(\frac{2-\sqrt{3}}{4}\right)^x+\left(\frac{2+\sqrt{3}}{4}\right)^x=1\)

TH1: \(x>1\)

Vì \(\frac{2+\sqrt{3}}{4};\frac{2-\sqrt{3}}{4}<1;x> 1 \Rightarrow \left ( \frac{2-\sqrt{3}}{4} \right )^x< \frac{2-\sqrt{3}}{4};\left ( \frac{2+\sqrt{3}}{4} \right )^x< \frac{2+\sqrt{3}}{4}\)

\(\Rightarrow \left ( \frac{2-\sqrt{3}}{4} \right )^x+\left ( \frac{2+\sqrt{3}}{4} \right )^x<\frac{2-\sqrt{3}}{4}+\frac{2+\sqrt{3}}{4}=1\) (vô lý)

TH2: \(x<1 \)

\(\frac{2+\sqrt{3}}{4};\frac{2-\sqrt{3}}{4}<1; x< 1 \Rightarrow \left ( \frac{2-\sqrt{3}}{4} \right )^x> \frac{2-\sqrt{3}}{4};\left ( \frac{2+\sqrt{3}}{4} \right )^x> \frac{2+\sqrt{3}}{4}\)

\(\Rightarrow \left ( \frac{2-\sqrt{3}}{4} \right )^x+\left ( \frac{2+\sqrt{3}}{4} \right )^x>\frac{2-\sqrt{3}}{4}+\frac{2+\sqrt{3}}{4}=1\) (vô lý)

Do đó \(x=1\)

a, 7\(x\) - \(x\) = 5²¹ : 5¹⁹ + 3.2².7

     6\(x\)    = 5³ + 3.4.7

    6\(x\)    = 125 + 12.7

    6\(x\)  = 125 + 84

    6\(x\) = 209

     \(x\)  = 209 : 6

    \(x\) = \(\dfrac{209}{6}\)

b; 11\(x\) - 7\(x\) + 3⁴ : 3³ = 5⁴ + 2\(x\)

    4\(x\) + 3 = 625 + 2\(x\)

   4\(x\) - 2\(x\) = 625 - 3

   2\(x\)        = 622

     \(x\)        = 622 : 2

    \(x\)        = 311

c; 75 - 5.(\(x-3\))³ = 700

          5.(\(x\) - 3)³ = 700 - 75

         5.(\(x\) - 3)³ = - 625

           (\(x\) - 30)³ = - 625 : 5

           (\(x\) - 30)³ = - 125

           (\(x-3\))³ =  (-5)³

           \(x\) - 3 = - 5 

           \(x\)         = - 5 + 3

            \(x\)       = -2

d, 3.(2\(x\) - 1)² = 75

       (2\(x\) - 1)² = 75 : 3

       (2\(x\) - 1)² = 25

       \(\left[{}\begin{matrix}2x-1=-5\\2x-1=5\end{matrix}\right.\)

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

       \(\left[{}\begin{matrix}2x=-4\\2x=6\end{matrix}\right.\)

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

\(x^{10}=x\Leftrightarrow x^{10}-x=0\Leftrightarrow x.\left(x^9-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\\ \)

`#040911`

`a,`

`15 + 25 \div (2x - 1) = 20`

`\Rightarrow 25 \div (2x - 1) = 20 - 15`

`\Rightarrow 25 \div (2x - 1) = 5`

`\Rightarrow 2x - 1 = 25 \div 5`

`\Rightarrow 2x - 1 = 5`

`\Rightarrow 2x = 6`

`\Rightarrow x = 3`

Vây, `x = 3.`

`b,`

\(3^{x-1}+2\cdot3^x=21\)

`\Rightarrow 3^x \div 3 + 2. 3^x = 21`

`\Rightarrow 3^x . \frac{1}{3} + 2. 3^x = 21`

`\Rightarrow 3^x . (\frac{1}{3} + 2) = 21`

`\Rightarrow 3^x . \frac{7}{3} = 21`

`\Rightarrow 3^x = 21 \div \frac{7}{3}`

`\Rightarrow 3^x = 9`

`\Rightarrow 3^x = 3^2`

`\Rightarrow x = 2`

Vậy, `x = 2.`

`c,`

\(2^{x-3}+2^{x+1}=17\)

`\Rightarrow 2^x \div 2^3 + 2^x . 2 = 17`

`\Rightarrow 2^x . \frac{1}{8} + 2^x . 2 = 17`

`\Rightarrow 2^x . (\frac{1}{8} + 2) = 17`

`\Rightarrow 2^x . \frac{17}{8} = 17`

`\Rightarrow 2^x = 17 \div \frac{17}{8}`

`\Rightarrow 2^x = 8`

`\Rightarrow 2^x = 2^3`

`\Rightarrow x = 3`

Vậy, `x = 3`

`d,`

\(5^x-5^{x-1}=20\)

`\Rightarrow 5^x - 5^x \div 5 = 20`

`\Rightarrow 5^x - 5^x . \frac{1}{5} = 20`

`\Rightarrow 5^x . (1 - \frac{1}{5} = 20`

`\Rightarrow 5^x . \frac{4}{5} = 20`

`\Rightarrow 5^x = 20 \div \frac{4}{5}`

`\Rightarrow 5^x = 25`

`\Rightarrow 5^x = 5^2`

`\Rightarrow x = 2`

Vậy, `x = 2.`

\(a.25:\left(2x-1\right)=5\)

\(2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)

\(b.3^x:3+2.3^x=21\)\(\Leftrightarrow3^x.\dfrac{1}{3}+2.3^x=21\)

\(\Leftrightarrow3^x\left(\dfrac{1}{3}+2\right)=21\)

\(\Leftrightarrow3^x.\dfrac{7}{3}=21\)

\(\Leftrightarrow3^x=9\Leftrightarrow x=2\)

\(c.2^x:2^3+2^x.2=17\Leftrightarrow2^x.\dfrac{1}{8}+2^x.2=17\)

\(\Leftrightarrow2^x.\dfrac{17}{8}=17\Leftrightarrow2^x=8\Leftrightarrow x=3\)

\(d.5^x-5^x:5=20\Leftrightarrow5^x-5^x.\dfrac{1}{5}=20\)

\(\Leftrightarrow5^x\left(1-\dfrac{1}{5}\right)=20\Leftrightarrow5^x=20:\dfrac{4}{5}\Leftrightarrow5^x=25\Leftrightarrow x=2\)

a) (9x+2).3=60

      9x+2    = 60:3

      9x+2    = 20

      9x        = 20-2

      9x        = 18

        x        = 2

c) \(44-7x=3^4:3^2\)

    \(44-7x=9\)

                \(7x=44-9\)

                \(7x=35\)

                   \(x=5\)

Hok tốt nha^^

thanks chúc bạn học tốt