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bài 1:
|x| = \(\dfrac{1}{3}\) => x = \(\pm\)\(\dfrac{1}{3}\) |y| = 1 => y = \(\pm\)1
a
+) A = 2x\(^2\) - 3x + 5
= 2\(\left(\dfrac{1}{3}\right)^2\) - 3.\(\dfrac{1}{3}\) +5 = 2.\(\dfrac{1}{9}\) - 1 + 5
= \(\dfrac{2}{9}\) - 1 + 5 = \(\dfrac{2-9+45}{9}\) = \(\dfrac{38}{9}\)
+) A = 2x\(^2\) - 3x + 5
= 2\(\left(\dfrac{-1}{3}\right)^2\) - 3\(\left(\dfrac{-1}{3}\right)\) + 5
= 2.\(\dfrac{1}{9}\) - (-1) + 5 = \(\dfrac{2}{9}\) + 1 +5
= \(\dfrac{2+9+45}{9}\) = \(\dfrac{56}{9}\)
b) +) B = 2x\(^2\) - 3xy + y\(^2\)
= 2\(\left(\dfrac{1}{3}\right)^2\) - 3.\(\dfrac{1}{3}\).1 + 1\(^2\)
= 2.\(\dfrac{1}{9}\) - 1 + 1 = \(\dfrac{2}{9}\) - 1 + 1
= \(\dfrac{2-9+9}{9}\) = \(\dfrac{2}{9}\)
+) B = 2x\(^2\) - 3xy + y\(^2\)
= 2\(\left(\dfrac{-1}{3}\right)\)\(^2\) - 3\(\left(\dfrac{-1}{3}\right)\). 1 + 1\(^2\)
= 2.\(\dfrac{1}{9}\) - (-1) + 1 = \(\dfrac{2}{9}\) + 1 + 1
= \(\dfrac{2+9+9}{9}\) = \(\dfrac{20}{9}\)
bài 3
x.y.z = 2 và x + y + z = 0
A = ( x + y )( y +z )( z + x )
= x + y . y + z . z + x = ( x + y + z ) + ( x . y . z )
= 0 + 2 = 2
bài 4
a) | 2x - \(\dfrac{1}{3}\) | - \(\dfrac{1}{3}\) = 0 => | 2x - \(\dfrac{1}{3}\) | = \(\dfrac{1}{3}\)
=> 2x - \(\dfrac{1}{3}\) = \(\pm\) \(\dfrac{1}{3}\)
+) 2x - \(\dfrac{1}{3}\)= \(\dfrac{1}{3}\)
=> 2x = \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)
x = \(\dfrac{2}{3}\) : 2 = \(\dfrac{2}{3}\) . \(\dfrac{1}{2}\) = \(\dfrac{1}{3}\)
+) 2x - \(\dfrac{1}{3}\) = \(\dfrac{-1}{3}\)
2x = \(\dfrac{-1}{3}\) + \(\dfrac{1}{3}\) = 0
x = 0 : 2 = 2
Bài 2: Tính giá trị của biểu thức:
a) P= 1/3 x^2 y + xy^2 - xy + 1/2 xy^2 - 5xy - 1/3 x^2 y (1)
Tại x = 0,5; y = 1
Thay \(x=0,5 ; y=1\) vào biểu thức (1) , ta có :
P= \(\dfrac{1}{3} . 0,5^2.1+0,5.1^2-0,5.1+\dfrac{1}{2}. 0,5.1^2-5.0,5.1-\dfrac{1}{3}.0,5^2.1\)
P= \(=\dfrac{1}{12}+\dfrac{1}{2} -0,5+\dfrac{1}{4} -\dfrac{5}{2} - \dfrac{1}{12}\)
P= \(= \dfrac{-9}{4}\)
Vậy \(P =\dfrac{-9}{4}\)
a) bn xem lại xem đề bài có đúng k nhé !
Nếu đúng thì kq sẽ là 1
b)
\(\Rightarrow x\in\begin{cases}0\\\frac{10}{3}\end{cases}\)
c)
\(+,x< -2\Rightarrow\left\{{}\begin{matrix}x+2< 0\\2x-3< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+2\right|=-2-x\\\left|2x-3\right|=3-2x\end{matrix}\right.\Rightarrow1-3x=5\Rightarrow x=-\frac{4}{3}\left(\text{loại}\right)\)
\(+,x\ge\frac{3}{2}\Rightarrow\left\{{}\begin{matrix}2x-3\ge0\\x+2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|2x-3\right|=2x-3\\\left|x+2\right|=x+2\end{matrix}\right.\Rightarrow3x-1=5\Rightarrow x=2\left(\text{thoa man}\right)\)
\(+,-2\le x< \frac{3}{2}\Rightarrow\left\{{}\begin{matrix}x+2\ge0\\2x-3< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+2\right|=x+2\\\left|2x-3\right|=3-2x\end{matrix}\right.\Rightarrow5-x=0\Rightarrow x=0\left(\text{thoa man}\right)\)
\(2.\text{ Ta co:}\left\{{}\begin{matrix}\left|x-102\right|\ge102-x\\\left|2-x\right|\ge x-2\end{matrix}\right.\Rightarrow A\ge102-x+x-2=100.\Rightarrow A_{min}=100.\text{dâu "=" xay ra}\Leftrightarrow\left\{{}\begin{matrix}102-x\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow2\le x\le102\)
Câu 1:
a: =>|2x+1|=5
=>2x+1=5 hoặc 2x+1=-5
=>2x=4 hoặc 2x=-6
=>x=2 hoặc x=-3
b: \(\Leftrightarrow\left[{}\begin{matrix}3x-2=x+1\\3x-2=-x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\4x=1\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{3}{2};\dfrac{1}{4}\right\}\)
Bài 2:
a: Ta có: ΔCAB cân tại C
mà CI là đường cao
nênI là trung điểm của AB
=>IA=IB
b: IA=IB=AB/2=6cm
\(IC=\sqrt{10^2-6^2}=8\left(cm\right)\)
Bài 1:
a) \(f\left(x\right)=2x\left(x^2-3\right)-4\left(1-2x\right)+x^2\left(x-1\right)+\left(5x+3\right)\)
\(=2x^3-6x-4+8x+x^3-x^2+5x+3\)
\(=x^3-x^2+7x-1\)
\(g\left(x\right)=-3\left(1-x^2\right)-2\left(x^2-2x+1\right)\)
\(=-3+3x^2-2x^2+4x-2\)
\(=x^2+4x-5\)
b) \(h\left(x\right)=f\left(x\right)-g\left(x\right)\)
\(=x^3-x^2+7x-1-x^2-4x+5\)
\(=x^3-2x^2+3x-4\)
a ) \(A=\left|x+1\right|+\left|x+2\right|-2x+3\ge2x+3-2x+3=6\)
Dấu " = " xảy ra khi \(\left(x+2\right)\left(x+1\right)\ge0\)
b )
\(B=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=4\)
Dấu " = " xảy ra khi \(\left(2x+3\right)\left(1-2x\right)\ge0\)
c )
\(C=\left|x-1\right|+\left|x-2\right|+\left|x-2\right|\ge\left|x-1\right|+\left|2-x\right|\ge\left|x-1+2-x\right|=1\)
Dấu " = " xảy ra khi \(x=2\)
câu a) mk k hiểu lắm!