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=> x(x - 1)=0
=> \(\left[\begin{array}{nghiempt}x=1\\x-1=0\end{array}\right.\)
=>\(\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
Vậy đa thức có nghiệm là x=0 ; x=1
1) \(x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-1=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
b) \(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\)
c)\(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=3\end{array}\right.\)
d)\(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\3x-4=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{4}{3}\end{array}\right.\)
bạn vào link này xem nhé
http://olm.vn/hoi-dap/question/97037.html
Do y tỉ lệ nghịch vs x theo hẹ số a = 12
=> y = \(\frac{12}{x}\)
a) y = \(\frac{12}{x}\)
+) f(-12) = \(\frac{12}{-12}\) = -1
+) f(-4) = \(\frac{12}{-4}=-3\)
+) f(3) = \(\frac{12}{3}=4\)
+) f(6) = \(\frac{12}{6}=2\)
b)
f(x)=4
\(\Leftrightarrow\) 12:x =4
\(\Leftrightarrow\) x =3
f(x) =0
\(\frac{12}{0}\) ( x ko xác định )
c)
\(\frac{12}{x}=\frac{12}{-x}\)
\(\frac{12}{x}=-\frac{12}{x}=\frac{12}{-x}\)
=> f(-x) = -f(x)
vậy \(\forall x\in R\) thì f(-x ) = -f(x)
c) -f(x) = \(\frac{-12}{x}\) (1)
f(-x)=\(\frac{12}{-x}=\frac{-12}{x}\) (2)
từ (1) và (2) => -f(x) = f(-x)
+ Với x < -5 thì |x + 5| = -(x + 5) = -x - 5
=> -x - 5 = 4x + 1
=> -x - 4x = 1 + 5
=> -5x = 6
=> \(x=-\frac{6}{5}\), không thỏa mãn x < -5
+ Với \(x\ge-5\) thì |x + 5| = x + 5
=> x + 5 = 4x + 1
=> 4x - x = 5 - 1
=> 3x = 4
=> \(x=\frac{4}{3}\), thỏa mãn \(x\ge-5\)
Vậy \(x=\frac{4}{3}\)
\(\left|x+5\right|=4x+1\)
\(=>\left[\begin{array}{nghiempt}x+5=4x+1\\x+5=-\left(4x+1\right)=-4x-1\end{array}\right.\)
\(=>\left[\begin{array}{nghiempt}3x=4\\5x=-6\end{array}\right.\)
\(=>\left[\begin{array}{nghiempt}x=\frac{4}{3}\\x=-\frac{6}{5}\end{array}\right.\)
giúp mình với. Mai mình phải nộp bài rồi.
4.
\(\left(0,36\right)^8=\left(\left(0,6\right)^2\right)^8=\left(0,6\right)^{16}\)
\(\left(0,216\right)^4=\left(\left(0,6\right)^3\right)^4=\left(0,6\right)^{12}\)
5.
a, \(\left(3\times5\right)^3=15^3=1125\)
b, \(\left(\frac{-4}{11}\right)^2=\frac{16}{121}\)
c, \(\left(0,5\right)^4\times6^4=\left(0,5\times6\right)^4=3^4=81\)
d, \(\left(\frac{-1}{3}\right)^5\div\left(\frac{1}{6}\right)^5=\left(\frac{-1}{3}\right)^5\times6^5=\left(\frac{-1}{3}\times6\right)^5=\left(-2\right)^5=-32\)
6.
a, \(\frac{6^2\times6^3}{3^5}=\frac{6^5}{3^5}=\frac{2^5\times3^5}{3^5}=2^5=32\)
b, \(\frac{25^2\times4^2}{5^5\times\left(-2\right)^5}=\frac{100^2}{\left(-10\right)^5}=\frac{10^4}{\left(-10\right)^5}=\frac{-1}{10}\)
c, Mình không nhìn rõ đề
d, \(\left(-2\frac{3}{4}+\frac{1}{2}\right)^2=\left(\frac{-11}{4}+\frac{1}{2}\right)^2=\left(\frac{-9}{4}\right)^2=\frac{81}{16}\)
7.
a, \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\Rightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\Rightarrow m=4\)
b, \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\Rightarrow\left(\frac{3}{5}\right)^n=\left(\left(\frac{3}{5}\right)^2\right)^5\Rightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\Rightarrow n=10\)
c, \(\left(-0,25\right)^p=\frac{1}{256}\Rightarrow\left(-0,25\right)^p=\left(\frac{1}{4}\right)^4\Rightarrow\left(-0,25\right)^p=\left(0,25\right)^4\Rightarrow p=4\)
8.
a, \(\left(\frac{2}{5}+\frac{3}{4}\right)^2=\left(\frac{23}{20}\right)^2=\frac{529}{400}\)
b, \(\left(\frac{5}{4}-\frac{1}{6}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
)
Vì \(f\left(x_1.x_2\right)=f\left(x_1\right).f\left(x_2\right)\) nên:
\(f\left(4\right)=f\left(2.2\right)=f\left(2\right).f\left(2\right)=10.10=100\)
\(f\left(16\right)=f\left(4.4\right)=f\left(4\right).f\left(4\right)=100.100=10000\)
\(f\left(32\right)=f\left(16.2\right)=f\left(16\right).f\left(2\right)=10000.10=100000\)
Vậy \(f\left(32\right)=100000\)
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
\(\Rightarrow x=2k\)
\(y=3k\)
\(z=5k\)
Thay \(x=2k;y=3k;z=5k\) vào \(x.y.z=810\) ta được:
\(2k.3k.5k=810\)
\(30k^3=810\)
\(k^3=27\)
\(k^3=3^3\)
\(\Rightarrow k=3\)
\(\Rightarrow x=2k=2.3=6\)
\(y=3k=3.3=9\)
\(z=5k=5.3=15\)
Vậy \(x=6;y=9;z=15\)
Ta có: \(\left|x-1\right|+\left|x-5\right|=\left|x-1\right|+\left|5-x\right|\)
Nhận thấy: \(\left[{}\begin{matrix}\left|x-1\right|\ge x-1\\\left|5-x\right|\ge5-x\end{matrix}\right.\)
\(\Rightarrow\left|x-1\right|+\left|5-x\right|\ge x-1+5-x\)
\(\Rightarrow\left|x-1\right|+\left|5-x\right|\ge4\)
Dấu \("="\) xảy ra khi:
\(\left[{}\begin{matrix}x-1\ge0\\5-x\ge0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge1\\x\le5\end{matrix}\right.\) \(\Rightarrow1\le x\le5\)
Vậy \(1\le x\le5.\)
Cho mk thêm cái ạ:
\(x\in\left\{1;2;3;4;5\right\}\)
Vậy \(x\in\left\{1;2;3;4;5\right\}\)
ta có:f(x)=4x2-5
f(-x)=4(-x)2-5=4x2-5
=> f(x)=f(-x)