Question

[Anh Triêt]

1. Tìm x biết:

a) \(\left(x+36\right)^2=1936\)

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

c) \(2005^{\left(2x-1\right)\left(2+3x\right)}=1\)

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

e) \(\left(\dfrac{3}{5}\right)^x.\left(\dfrac{125}{72}\right)^5=\dfrac{81}{625}\)

2. a) Tìm x biết:

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

b) Có số hữu tỉ x thõa mản đẳng thức

\(\left(3x+1\right)^4+\left(3x-1\right)^2=0\)

3. Tính giá trị của biểu thức:

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

4. Tìm các cặp số (x,y) thõa mản các đk sau:

a) \(x\left(x+y\right)=\dfrac{5}{12}\) và \(y\left(x+y\right)=\dfrac{5}{18}\)

b) \(x\left(x-y\right)=\dfrac{2}{45}\) và \(y\left(x-y\right)=\dfrac{1}{25}\)

Mai nộp rồi còn 4 bài đó ( help me )

Answer 1

[Ngô Thanh Sang]

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

Answer 2

[Ngô Thanh Sang]

3. Tính giá trị của biểu thức:

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

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

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