Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Trả lời:
\(B=\left(x-3\right).\left(x+3\right).\left(x^2+9\right)-\left(x^2+2\right).\left(x^2-2\right)\)
\(B=\left(x^2-9\right).\left(x^2+9\right)-\left(x^4-4\right)\)
\(B=\left(x^4-81\right)-\left(x^4-4\right)\)
\(B=x^4-81-x^4+4\)
\(B=-77\)
\(A=\left|4x-3\right|+\left|5y+7,5\right|+10\)
Mà \(\left|4x-3\right|\ge0\)với mọi x
\(\left|5y+7,5\right|\ge0\)với mọi y
\(\Rightarrow A\)có GTNN là 10
Để A có GTNN thì :
\(4x-3=0\) \(5y+7,5=0\)
\(4x=3\) \(5y=-7,5\)
\(x=\frac{3}{4}\) \(y=-1,5\)
\(B=\frac{5,8}{\left|2,5-x\right|+5,8}\)
Mà \(\left|2,5-x\right|\ge0\)
\(\Rightarrow\)GTNN \(\left|2,5-x\right|+5,8=5,8\)
Để B có GTLN \(\Rightarrow2,5-x=0\)
\(\Rightarrow x=2,5\)
a)(x − 12)2 = 0
=>x − 12 = 0
=> x = 12
b) (x+12)2 = 0,25
=> x + 12 = 0,5 hoặc x + 12= -0,5
=> x = -11,5 hoặc x = -12,5
c) (2x−3)3 = -8
=> 2x - 3 = -2
=> x = 0,5
d) (3x−2)5 = −243
=> 3x - 2 = -3
=> x = -1/3
e) (7x+2)-1 = 3-2
=> \(\dfrac{1}{7x+2}=\dfrac{1}{9}\)
=> 7x + 2 = 9
=> x = 1
f) (x−1)3 = −125
=> (x−1) = −5
=> x = -4
g) (2x−1)4 = 81
=> 2x - 1 = 3
=> x = 2
h) (2x−1)6 = (2x−1)8
=> 2x -1 = 0 hoặc 2x - 1 = 1 hoặc 2x - 1 = -1
=> x = 1/2 hoặc x = 1 hoặc x = 0
a/ \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
b/ \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^2\\\left(x+\dfrac{1}{2}\right)^2=\left(-\dfrac{1}{2}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{2}\\x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy ..
c/ \(\left(2x-3\right)^3=-8\)
\(\Leftrightarrow\left(2x-3\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-3=-2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
d/ \(\left(3x-2\right)^5=-243\)
\(\left(3x-2\right)^5=\left(-3\right)^5\)
\(\Leftrightarrow3x-2=-3\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy ...
e/ \(\left(x-1\right)^3=-125\)
\(\Leftrightarrow\left(x-1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x-1=-5\)
\(\Leftrightarrow x=-4\)
Vậy..
f/ \(\left(2x-1\right)^4=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=3^4\\\left(2x-1\right)^4=\left(-3\right)^4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy...
g/ \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy..
Bài 2, \(\left(x-1\right)^3=27\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
Bài 3, \(-2,4-\frac{2}{3}< x\le\frac{5}{3}-1\frac{2}{5}\)
\(\Leftrightarrow-3,0\left(6\right)< x\le0,2\left(6\right)\)
Vì x nguyên nên \(x\in\left\{-3;-2;-1;0\right\}\)
Bài 4, Từ \(2x=3y=4z\)
\(\Rightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\)(cùng chia cho 12)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}=\frac{x+y+z}{6+4+3}=\frac{130}{13}=10\)
\(\Rightarrow\hept{\begin{cases}x=6.10=60\\y=4.10=40\\z=3.10=30\end{cases}}\)
A=(x-2)2 + \(\sqrt{ }\)3 .Ta có:(x-2)2 \(\ge\) 0 suy ra (x+2)2+ căn 3\(\ge\)căn 3 .Dấu = xảy ra \(\Leftrightarrow\)x=-2.Vạy Amin=căn 3\(\Leftrightarrow\)x=-2
B mik nghĩ là sai đề
C làm tương tự câu a
Ta có: \(x^2=\left|x^2\right|\)
Áp dụng BĐT |a| + |b| \(\ge\left|a+b\right|\),ta được:
\(\left|x+1\right|+x^2\)\(=\left|x+1\right|+\left|x^2\right|\ge\left|x^2+x+1\right|\)
\(=\left|\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right|=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)(Vì \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\))
(Dấu "="\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\))
Lại có: \(\left(2y-1\right)^2\ge0\)(Dấu "="\(\Leftrightarrow2y-1=0\Leftrightarrow y=\frac{1}{2}\))
Vậy \(C\ge\frac{3}{4}+0+3=\frac{15}{4}\)
(Dấu "="\(\Leftrightarrow x=-\frac{1}{2}\);\(y=\frac{1}{2}\))
Gioi qua troi