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\(\frac{1}{9}.3^4.3^x=3^7\)
\(\Leftrightarrow3^x=3^7:\frac{1}{9}:3^4=243\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Thêm nữa câu a) Tính: M(x) + N(x)+ P(x)
B) Tính M(x) - N (x) - P(x)
ok rồi giúp mình với nha
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\(f\left(x\right)=-3x^2+x-1+x^4-x^3-x^2+3x^4+2x^3\)
\(f\left(x\right)=\left(x^4+3x^4\right)-\left(x^3-2x^3\right)-\left(3x^2+x^2\right)+x-1\)
\(f\left(x\right)=4x^4+x^3-4x^2+x-1\)
\(g\left(x\right)=x^4+x^2-x^3+x-5+5x^3-x^2-3x^4\)
\(g\left(x\right)=\left(x^4-3x^4\right)+\left(5x^3-x^3\right)+\left(x^2-x^2\right)+x-5\)
\(g\left(x\right)=-2x^4+4x^3+x-5\)
`@` `\text {Ans}`
`\downarrow`
`a,`
\(f(x) -3x^2 + x - 1 + x^4 - x^3 - x^2 + 3x^4 + 2x^3\)
`= (x^4 +3x^4) + (-x^3 +2x^3) + (-3x^2 - x^2) + x - 1`
`= 4x^4 + x^3 -4x^2 + x -1`
\(g(x) = x^4 + x^2 - x^3 + x - 5 + 5x^3 - x^2 - 3x^4\)
`= (x^4-3x^4) + (-x^3+5x^3) + (x^2 - x^2) + x -5`
`= -2x^4 + 4x^3 +x - 5`
![](https://rs.olm.vn/images/avt/0.png?1311)
Vũ Hồng Linh bạn check lại bài đầu dùm =_="
\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)
\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)
\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)
\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)
=> Làm nốt
Mấy bài kia cũng làm tương tự
(- \(\dfrac{1}{3}\))3.\(x\) = \(\dfrac{1}{81}\)
\(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))3
\(x\) = - (\(\dfrac{1}{3}\))4 :(\(\dfrac{1}{3}\))3
\(x=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 8:
a: \(\left(\dfrac{2}{5}+\dfrac{3}{4}\right)^2=\left(\dfrac{8+15}{20}\right)^2=\left(\dfrac{23}{20}\right)^2=\dfrac{529}{400}\)
b: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2^3:\left|x-2\right|=2\)
\(\Leftrightarrow8:\left|x-2\right|=2\)
\(\Leftrightarrow\left|x-2\right|=8:2\)
\(\Leftrightarrow\left|x-2\right|=4\)
Xét trường hợp 1: \(x-2=4\)
\(\Rightarrow x=4+2\)
\(\Rightarrow x=6\)
Xét trường hợp 2: \(x-2=-4\)
\(\Rightarrow x=-4+2\)
\(\Rightarrow x=-\left(4-2\right)\)
\(\Rightarrow x=-2\)
Vậy \(x=6\) hoặc \(x=-2\)
b)
1- X2= 3/4
Ta có: 1- X2= 3/4
X2= 1- 3/4
X2= 1/4
=) X2= (1/2 )2 hoặc X2= (-1/2 ) 2
Vậy, Xϵ { 1/2 ; -1/2}
1-x2=3/4
x2=1-3/4
x2=1/4
x=1/2