a: Đặt f(x)=0

\(\Leftrightarrow x-2x^2+2x^2-x+4=0\)

=>4=0(loại)

b: Đặt g(x)=0

\(\Leftrightarrow x^2-5x-x^2-2x+7x=0\)

=>0x=0(luôn đúng)

c: Đặt H(x)=0

\(\Leftrightarrow x^2-x+1=0\)

Δ=1-4=-3<0

DO đó: PTVN

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k tui và kb nhé

tạm biệt các bạn

a) x = 2 hoặc x = -2

b) x 1 hoặc x = -1

- Ủng hộ -

1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)

=>4x=18

hay x=9/2

2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)

=>4x=108

hay x=27

3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)

\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)

=>4x=12

hay x=3

a) \(\left(x-5\right)\left(x+8\right)-\left(x+4\right)\left(x-1\right)\)

\(=\left(x^2+3x-40\right)-\left(x^2+3x-4\right)\)

\(=x^2+3x-40-x^2-3x+4\)

\(=-36\)

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

\(=x^4\left(x^4-1\right)\)

\(=x^8-x^4\)

1. Tìm x:

a) \(\left(x+36\right)^2=1936\Leftrightarrow x+36=\pm44.\) Vậy x = 8 hoặc x = -80

b) \(\left(\dfrac{3}{5}\right)^{x+2}=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}\right)^{x+2}=\left(\dfrac{3}{5}\right)^4\Leftrightarrow x+2=4\Leftrightarrow x=2\)

c) Xem lại đề

d) \(\left(\dfrac{9}{16}\right)^{x-5}=\left(\dfrac{4}{3}\right)^4\Leftrightarrow\left(\dfrac{3}{4}\right)^{2\left(x-5\right)}=\left(\dfrac{3}{4}\right)^{-4}\Leftrightarrow2\left(x-5\right)=-4\Leftrightarrow x=3\)

e) \(\left(\dfrac{3}{5}\right)^x.\left(\dfrac{125}{27}\right)^x=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}.\dfrac{125}{27}\right)^x=\left(\dfrac{3}{5}\right)^4\Leftrightarrow\left(\dfrac{5}{3}\right)^{2x}=\left(\dfrac{5}{3}\right)^{-4}\Leftrightarrow2x=-4\) Vậy x = -2

3. Tính giá trị của biểu thức:

\(A=\left\{-\left[\left(\dfrac{1}{x}\right)^2\right]^3\right\}^5.\left\{-\left[\left(-x\right)^5\right]^2\right\}^3\) \(\left(x\notin0\right)\)

\(=\left\{-\left[-\dfrac{1}{x^2}\right]^3\right\}^5.\left\{-\left[-\left(-x\right)^5\right]^2\right\}^3=\left\{-\left[-\dfrac{1}{x^6}\right]\right\}^5.\left\{-\left[x^5\right]^2\right\}^3\)

\(=\left\{\dfrac{1}{x^6}\right\}^5.\left\{-x^{10}\right\}^3=\dfrac{1}{x^{30}}.\left(-x^{30}\right)=-1\)

a, |- \(x\) + 2|  - |\(x\) + 7| = 0

           |- \(x\) + 2| = | \(x\) + 7|

           \(\left[{}\begin{matrix}-x+2=x+7\\-x+2=-x-7\end{matrix}\right.\)         

            \(\left[{}\begin{matrix}x=-\dfrac{5}{2}\\2=-7\left(loại\right)\end{matrix}\right.\)

              vậy \(x\) = -\(\dfrac{5}{2}\)

b, |2\(x\) - 1| + |2 + y| ≥ 0

     |2\(x\) - 1| ≥ 0 ∀ \(x\)

     |2 + y| ≥ 0 ∀ y 

  ⇒ |2\(x\) - 1| +|2 + y| ≥ 0 ∀\(x\) ; y

làm đc câu c thôi à dc ko bạn