tìm nghiệm ạaa
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
\(a,\dfrac{-1}{3}.\dfrac{5}{7}=\dfrac{-5}{21}\\ b,\dfrac{1}{2}.\dfrac{-3}{4}=\dfrac{-3}{8}\\ c,\dfrac{19}{7}.\dfrac{7}{15}=\dfrac{19.1}{1.15}=\dfrac{19}{15}\\ d,\dfrac{5}{11}:\left(-9\right)=\dfrac{5}{11}.\dfrac{-1}{9}=\dfrac{-5}{99}\\ e,-1:\dfrac{3}{5}=-1.\dfrac{5}{3}=-\dfrac{5}{3}\\ f,\dfrac{-2}{9}:\dfrac{2}{9}:\dfrac{-9}{14}=-\left(\dfrac{2}{9}:\dfrac{2}{9}\right).\dfrac{-14}{9}=-1.\dfrac{-14}{9}=\dfrac{14}{9}\\ k,\left(-\dfrac{2}{7}\right)^2=\left(-1\right)^2.\left(\dfrac{2}{7}\right)^2=1.\dfrac{2^2}{7^2}=\dfrac{4}{49}\\ l,\dfrac{1}{3}.\left(\dfrac{1}{3}\right)^3=\left(\dfrac{1}{3}\right)^4=\dfrac{1^4}{3^4}=\dfrac{1}{243}\\ m,\left(-\dfrac{1}{2}\right)^2.\left(\dfrac{1}{2}\right)^3=\left(-1\right)^2.\left(\dfrac{1}{2}\right)^{2+3}=1.\left(\dfrac{1}{2}\right)^5=1.\dfrac{1^5}{2^5}=\dfrac{1}{32}\)
Bài 1:
\(a,\dfrac{1}{-8}+\dfrac{-5}{8}=\dfrac{-1}{8}+\dfrac{-5}{8}=\dfrac{-\left(1+5\right)}{8}=-\dfrac{6}{8}=\dfrac{-6:2}{8:2}=-\dfrac{3}{4}\\ b,\dfrac{1}{7}+\dfrac{-3}{7}=\dfrac{1-3}{7}=-\dfrac{2}{7}\\ c,\dfrac{-12}{35}+\dfrac{-7}{35}=\dfrac{-\left(12+7\right)}{35}=\dfrac{-19}{35}\\ d,\dfrac{1}{6}+\dfrac{2}{5}=\dfrac{1.5+2.6}{6.5}=\dfrac{5+12}{30}=\dfrac{17}{30}\\ e,\dfrac{3}{5}+\dfrac{-7}{4}=\dfrac{3.4-7.5}{5.4}=\dfrac{12-35}{20}=\dfrac{-23}{20}\\ g,-2-\left(-\dfrac{1}{5}\right)=-2+\dfrac{1}{5}=\dfrac{-2.5+1}{5}=\dfrac{-9}{5}\\ h,\dfrac{2}{3}-\left(-1\right)=\dfrac{2}{3}+1=\dfrac{5}{3}\\ i,4-\dfrac{2}{3}=\dfrac{4.3-2}{3}=\dfrac{12-2}{3}=\dfrac{10}{3}\\ j,\dfrac{3}{4}-2=\dfrac{3-2.4}{4}=\dfrac{-5}{4}\\ k,-1-\left(-\dfrac{2}{3}\right)=-1+\dfrac{2}{3}=\dfrac{-1.3+2}{3}=\dfrac{-3+2}{3}=-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bán kính là: 125,6/3,14/2=20 cm
S là:
20x20x3,14=1256 cm2
![](https://rs.olm.vn/images/avt/0.png?1311)
Var a:array:[1..100] of integer;
i,max:integer;
Begin
For i:=1 to 100 do
Begin
Write('Nhap so thu ',i,' = ');readln(a[i]);
End;
max:=a[1];
For i:=2 to n do
If a[i] > max then max:=a[i];
write('So lon nhat la ',max);
Readln
End.
Uses crt;
var i,n,max; longint;
a: array[1..100] of longint;
begin clrscr;
readln(n);
for i:=1 to n do read(n); readln;
max:=a[1];
for i:=1 to n do if(max<a[i]) then max:=a[i];
write('So nguyen lon nhat la: ',max);
readln;
end.
![](https://rs.olm.vn/images/avt/0.png?1311)
Quan hệ từ: ''Vì '' Tham khảo: ý nghĩa : những quan hệ từ trên giúp nối 2 câu đơn thành câu ghép và đồng thời thể hiện được sự hi sinh, tình cảm mà người bà dành cho cháu và tình cảm của cháu dành cho người bà
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có \(y'=\left(x^6\left(1-x\right)^5\right)'\)
\(=\left(x^6\right)'\left(1-x\right)^5+\left[\left(1-x\right)^5\right]'.x^6\)
\(=6x^5\left(1-x\right)^5+5\left(1-x\right)^4\left(1-x\right)'.x^6\)
\(=6x^5\left(1-x\right)^5-5x^6\left(1-x\right)^4\)
\(=x^5\left(1-x\right)^4\left[6\left(1-x\right)-5x\right]\)
\(=x^5\left(1-x\right)^4\left(6-11x\right)\)
\(y'=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{6}{11}\end{matrix}\right.\)
Vậy hàm số đã cho đạt cực trị tại \(x=0,x=1,x=\dfrac{11}{6}\)
2) Có \(y'=-2.\left(2x\right)'\sin2x\) \(=-4\sin2x\)
\(y'=0\Leftrightarrow\sin2x=0\) \(\Leftrightarrow2x=k\pi\left(k\inℤ\right)\) \(\Leftrightarrow x=\dfrac{k\pi}{2}\) \(\left(k\inℤ\right)\)
Vậy hàm số đã cho đạt cực trị tại \(x=\dfrac{k\pi}{2}\left(k\inℤ\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
X x ( 4,2 + 0,8 ) = 2,64
X x 5 = 2,64
X = 2,64 : 5
X = 0,528
X x ( 4,2 + 0,8 ) = 2,64
X x 5 = 2,64
X = 2,64 : 5
X = 0,528 ok
Đặt \(M\left(x\right)=0\Rightarrow3.\left(2-x\right)+\left(x-2\right)^2=0\)
\(\Rightarrow-3.\left(x-2\right)+\left(x-2\right)^2=0\)
\(\Rightarrow\left(x-2\right).\left[\left(-3\right)+\left(x-2\right)\right]=0\\ \Rightarrow\left(x-2\right).\left(x-2-3\right)=0\\ \Rightarrow\left(x-2\right).\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy \(M\left(x\right)\) có nghiệm là \(x=2,x=5\).
`M(x)=3.(2-x)+(x-2)^2`
Ta có :
`3.(2-x)+(x-2)^2=0`
`=>3.(2-x)+(2-x)^2=0`
`=>(2-x).(3+2-x)=0`
`=>(2-x).(5-x)=0`
\(\Rightarrow\left[{}\begin{matrix}2-x=0\\5-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)