Tìm X :
a) (1780-973) x (75:X) =2401+20
b)13x X-3xX-56=44
c) x/10=2/5
d) (1+4+7+.....+100) :X=17
Tính giá trị biểu thức:
A=62xm+33xn-2xm-3xn biết 2xm+n=15
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a) (y – 1) : 105 = 125 x 80
(y – 1) : 105 = 10000
(y – 1) = 10000 x 105
(y – 1) = 1050000
y = 1050000 + 1
y = 1050001
b) (y – 607 200) : 305 = 642 + 318
(y – 607 200) : 305 = 960
(y – 607 200) = 960 x 305
(y – 607 200) = 292800
y = 292800 + 607 200
y = 900000
\(a.\left(y-1\right):105=125x80\) \(b.\left(y-607200\right):305=642+318\)
\(\left(y-1\right):105=10000\) \(\left(y-607200\right):305=960\)
\(y-1=10000x105=1050000\) \(y=960+607200=608160\)
\(y=1050000+1=1050001\)
\(c.\left(1780-973\right)x\left(75:y\right)=2401+20\)
\(807x\left(75:y\right)=2401+20\)
\(807x\left(75:y\right)=2421\)
\(75:y=2421:807=3\)
\(y=75:3=25\)
a.
Phương trình hoành độ giao điểm:
\(x^2+6x+3=-2mx-m^2\Leftrightarrow x^2+2\left(m+3\right)x+m^2+3=0\)
\(\Delta'=\left(m+3\right)^2-\left(m^2+3\right)=6\left(m+1\right)>0\Rightarrow m>-1\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_A+x_B=-2\left(m+3\right)\\x_Ax_B=m^2+3\end{matrix}\right.\)
\(P=10\left(m+3\right)-2\left(m^2+3\right)=-2m^2+10m+24\)
\(P=-2\left(m-\dfrac{5}{2}\right)^2+\dfrac{73}{2}\le\dfrac{73}{2}\)
\(P_{max}=\dfrac{73}{2}\) khi \(m=\dfrac{5}{2}\)
b.
Pt hoành độ giao điểm:
\(x^2-2x-2=x+m\Leftrightarrow x^2-3x-m-2=0\)
\(\Delta=9+4\left(m+2\right)>0\Rightarrow m>-\dfrac{17}{4}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_A+x_B=3\\x_Ax_B=-m-2\end{matrix}\right.\)
Đồng thời \(y_A=x_A+m\) ; \(y_B=x_B+m\)
\(P=OA^2+OB^2=x_A^2+y_A^2+x_B^2+y_B^2\)
\(=x_A^2+x_B^2+\left(x_A+m\right)^2+\left(x_B+m\right)^2\)
\(=2\left(x_A^2+x_B^2\right)+2m\left(x_A+x_B\right)+2m^2\)
\(=2\left(x_A+x_B\right)^2-4x_Ax_B+2m\left(x_A+x_B\right)+2m^2\)
\(=18-4\left(-m-2\right)+6m+2m^2\)
\(=2m^2+10m+26=2\left(m+\dfrac{5}{2}\right)^2+\dfrac{27}{2}\ge\dfrac{27}{2}\)
Dấu "=" xảy ra khi \(m=-\dfrac{5}{2}\)
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`
`b)` Với `x ne -1;x ne -5` có:
`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`
`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`
`A=[x^2-3x-4]/[(x+1)(x+5)]`
`A=[(x+1)(x-4)]/[(x+1)(x+5)]`
`A=[x-4]/[x+5]`
`c)` Với `x ne -5; x ne -1; x ne 4` có:
`P=A.B=[x-4]/[x+5].[-10]/[x-4]`
`=[-10]/[x+5]`
Để `P` nguyên `<=>[-10]/[x+5] in ZZ`
`=>x+5 in Ư_{-10}`
Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`
`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)
\(A=\left(x-1\right)^2+8\ge8\\ A_{min}=8\Leftrightarrow x=1\\ B=\left(x+3\right)^2-12\ge-12\\ B_{min}=-12\Leftrightarrow x=-3\\ C=x^2-4x+3+9=\left(x-2\right)^2+8\ge8\\ C_{min}=8\Leftrightarrow x=2\\ E=-\left(x+2\right)^2+11\le11\\ E_{max}=11\Leftrightarrow x=-2\\ F=9-4x^2\le9\\ F_{max}=9\Leftrightarrow x=0\)
BÀI 1:
Ta có: \(VT=\left(7x+1\right)^2-\left(x+7\right)^2\)
\(=\left(7x+1+x+7\right)\left(7x+1-x-7\right)\)
\(=\left(8x+8\right)\left(6x-6\right)\)
\(=8\left(x+1\right).6\left(x-1\right)\)
\(=48\left(x^2-1\right)=VP\) (đpcm)
Bài 2:
\(16x^2-\left(4x-5\right)^2=15\)
\(\Leftrightarrow\)\(16x^2-16x^2+40x-25=15\)
\(\Leftrightarrow\)\(40x=40\)
\(\Leftrightarrow\)\(x=1\)
Vậy...
Bài 3:
\(A=x^2+2x+3=\left(x+1\right)^2+2\ge2\)
Vậy MIN A = 2 khi x = -1