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a.\(\left(3x-2\right)^2=16\)
Ta có: \(\left(3x-2\right)^2=16\)
\(\Rightarrow\left(3x-2\right)^2=\left(4\right)^2\)
\(\Rightarrow3x-2=4\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
b. \(\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\dfrac{-8}{125}\)
\(\Rightarrow\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\left(\dfrac{-2}{5}\right)^3\)
\(\Rightarrow\dfrac{4}{5}x-\dfrac{3}{4}=\dfrac{-2}{5}^{ }\)
\(\Rightarrow\dfrac{4}{5}x-=\dfrac{7}{20}\)
\(\Rightarrow x=\dfrac{7}{16}\)
a)\(-\left(\frac{-1}{2}xy^2z\right)^2\left(4x^2yz^3\right)\)
\(=-\left(\frac{1}{4}x^2y^4z^2\right)\left(4x^2yz^3\right)\)
\(=\left(\frac{-1}{4}.4\right)\left(x^2x^2\right)\left(y^4y\right)\left(z^2z^3\right)\)
\(=-x^4y^5z^5\) \(\Rightarrow\)Bậc là 14 Hệ số là -1
b)\(\left(\frac{-1}{3}x^2yz^3\right).\left(\frac{-6}{7}xyz^2\right)\)
\(=\left(\frac{-1}{3}.\frac{-6}{7}\right)\left(x^2x\right)\left(yy\right)\left(z^3z^2\right)\)
\(=\frac{2}{7}x^3y^2z^5\) \(\Rightarrow\)Bậc là 10 Hệ số là \(\frac{2}{7}\)
c)\(-3x^2.y^4.\left(\frac{-1}{3}y^4z^5x\right).\left(\frac{-1}{2}zyx^3\right)\)
\(=\left(-3.\frac{-1}{3}.\frac{-1}{3}\right)\left(x^2xx^3\right)\left(y^4y^4y\right)\left(z^5z\right)\)
\(=\frac{-1}{3}x^6y^9z^6\) \(\Rightarrow\)Bậc là 21 Hệ số là \(\frac{-1}{3}\)
d)\(\frac{3}{4}xy^3\left(\frac{-2}{3}x^2y^4\right)^2\)
\(=\frac{3}{4}xy^3\left(\frac{4}{9}x^4y^{16}\right)\)
\(=\left(\frac{3}{4}\cdot\frac{4}{9}\right)\left(xx^4\right)\left(y^3y^{16}\right)\)
\(=\frac{1}{3}x^5y^{19}\)
a/ \(\left|2x-1,6\right|-2,3=1,4\)
\(\Leftrightarrow\left|2x-1,6\right|=3,7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5,3\\2x=-2,1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)
Vậy ....
b/ \(5,4-\left|3x-1,2\right|=5,5\)
\(\Leftrightarrow\left|3x-1,2\right|=-0,1\)
Mà \(\left|3x-1,2\right|\ge0\)
\(\Leftrightarrow x\in\varnothing\)
c/ \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+1,3+x+2,4=4x\)
\(\Leftrightarrow2x+3,7=4x\)
\(\Leftrightarrow3,7=4x-2x\)
\(\Leftrightarrow2x=3,7\)
\(\Leftrightarrow x=1,85\)
Vậy ....
d/ \(\left|x-1,2\right|+\left|2,5-x\right|=0\)
Mà \(\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\) (loại)
Vậy ..
a, \(\left|2x-1,6\right|-2,3=1,4\)
\(\Rightarrow\left|2x-1,6\right|=3,7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)
b,\(5,4-\left|3x-1,2\right|=5,5\)
\(\Rightarrow\left|3x-1,2\right|=-0,1\) (vô lí)
Vì \(\left|x\right|\ge0\) mà \(\left|3x-1,2\right|< 0\)
Vậy, không có giá trị của x thỏa mãn.
c, \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+1,3+x+2,4=4x\)
\(\Leftrightarrow x+x+1,3+2,4=4x\)
\(\Leftrightarrow2x+3,7=4x\)
\(\Leftrightarrow2x-4x=-3,7\)
\(\Leftrightarrow-2x=-3,7\)
\(\Leftrightarrow x=\dfrac{3,7}{2}\)
d, \(\left|x-1,2\right|+\left|2,5-x\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\)
a) Ta có: \(-2xy^2\cdot\left(x^3y-2x^2y^2+5xy^3\right)\)
\(=-2x^4y^3+4x^3y^4-10x^2y^5\)
b) Ta có: \(\left(-2x\right)\cdot\left(x^3-3x^2-x+1\right)\)
\(=-2x^4+6x^3+2x^2-2x\)
c) Ta có: \(3x^2\left(2x^3-x+5\right)\)
\(=6x^5-3x^3+15x^2\)
d) Ta có: \(\left(-10x^3+\frac{2}{5}y-\frac{1}{3}z\right)\cdot\left(-\frac{1}{2}xy\right)\)
\(=5x^4y-\frac{1}{5}xy^2+\frac{1}{6}xyz\)
e) Ta có: \(\left(3x^2y-6xy+9x\right)\cdot\left(-\frac{4}{3}xy\right)\)
\(=-4x^3y^2+8x^2y^2-12x^2y\)
f) Ta có: \(\left(4xy+3y-5x\right)\cdot x^2y\)
\(=4x^3y^2+3x^2y^2-5x^3y\)
a/ \(M=x^4-xy^3+x^3y-y^4-1\)
\(\Leftrightarrow M=x^3\left(x+y\right)-y^3\left(x+y\right)-1\)
Mà \(x+y=0\)
\(\Leftrightarrow M=x^3.0-y^3.0-1\)
\(\Leftrightarrow M=-1\)
Vậy ...
\(C=5x^2+20x+2010\)
\(=5\left(x^2+4x+402\right)\)
\(=5\left(x^2+2.x.2+2^2+398\right)\)
\(=5\left[\left(x+2\right)^2+398\right]\)
VÌ \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+398\ge398\forall x\)
\(\Rightarrow C=5\left[\left(x+2\right)^2+398\right]\ge1990\forall x\)
Dấu "=" xảy ra <=> x = - 2
Vậy gtnn của C là 1990 tại x = - 2
1) Giá trị nhỏ nhất của A = 0
2) Giá trị nhỏ nhất của B = 2011
3) Gí trị nhỏ nhất của C = 2010
nếu bạn cần cách giải chi tiết thì nhắn tin gửi cho mk; mk sẽ giải cho
\(C=\frac{7}{9}x^3y^2\left(\frac{6}{11}axy^3\right)+\left(-5bx^2y^4\right)\left(\frac{-1}{2}axz\right)+ax\left(x^2y\right)^3\)
\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax\left(x^6y^3\right)\)
\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax^7y^3\)
\(D=\frac{\left(3x^4y^4\right)^2\left(\frac{6}{11}x^3y\right)\left(8x^{n-7}\right)\left(-2x^{7-n}\right)}{15x^3y^2\left(0,4ax^2y^2z^2\right)^2}\)
\(D=\frac{\left[3.\frac{6}{11}.8.\left(-2\right)\right]\left(x^8x^3x^{n-7}x^{7-n}\right)\left(y^8y\right)}{15.0,4.\left(x^3x^4\right)\left(y^2y^4\right)z^4a}\)
\(D=\frac{\frac{-188}{11}x^{24}y^9}{6x^7y^6z^4a}\)
a) Ta có: \(5x^2-3x\left(x+2\right)\)
\(=5x^2-3x^2-6x\)
\(=2x^2-6x\)
b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)
\(=3x^2-15x-5x^2-35x\)
\(=-2x^2-50x\)
c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)
\(=3x^2y\left(2x^2-y\right)-2x^2y\left(2x^2-y\right)\)
\(=x^2y\left(2x^2-y\right)=2x^4y-x^2y^2\)
d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)
\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)
\(=6x^2y-3x^2-10x^2y+6x^2+2x^2-2x\)
\(=-4x^2y+5x^2-2x\)
e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)
\(=4x^4-16x^3+4x^4-2x^3+14x^2\)
\(=8x^4-18x^3+14x^2\)
f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)
\(=25x-12x+4+35x-14x^3\)
\(=-14x^3+48x+4\)
a)
\(\left\{{}\begin{matrix}\left(4x-1\right)^4\ge0\\\left|2x-3y\right|\ge0\end{matrix}\right.\) \(\Rightarrow A\ge25,6\) tự tìm cận
không có Max
b) giống vậy
c) \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\Rightarrow-\left(x-3\right)^2\le0\\\left|4x-3y\right|\ge0\Rightarrow-\left|4x-3y\right|\le0\end{matrix}\right.\)
\(C\le40,5\) tự tìm cận
không có GTNN