Mọi người giải giùm ạ
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TT
2
TT
2
DT
Đỗ Thanh Hải
CTVVIP
25 tháng 5 2021
Đề dài thế này sao giải thích nhanh cho e đc
Part 1
1 C
2 B
3 D
4 C
5 B
6 A
Part 2
1 T
2 F
3 F
4 F
V
1 That old house has just been bought
2 If he doesn't take these pills, he won't be better
3 I suggest taking a train
4 Spending the weekend in the countryside is very wonderful
A
1
TT
18 tháng 6 2023
\(\left\{{}\begin{matrix}x+y=7\\-x+2y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x-y=-7\\-x+2y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=7\\\left[-x-\left(-x\right)\right]+\left(-y-2y\right)=-7-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=7\\-3y=-9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=7\\y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3=7\\y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\)
Vậy hệ pt có nghiệm duy nhất \(\left(x;y\right)=\left(4;3\right)\)
DD
1 tháng 11 2023
Bài `13`
\(a,\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}\\ =\sqrt{9\cdot3}+\sqrt{16\cdot3}-\sqrt{36\cdot3}-\sqrt{4\cdot3}\\ =3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}\\ =\left(3+4-6-2\right)\sqrt{3}\\ =-\sqrt{3}\\ b,\left(\sqrt{28}+\sqrt{12}-\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{84}\\ =\left(\sqrt{4\cdot7}+\sqrt{4\cdot3}-\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{4\cdot21}\\ =\left(2\sqrt{7}+2\sqrt{3}-\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\\ =2\cdot7+2\sqrt{21}-7+2\sqrt{21}\\ =14+2\sqrt{21}-7+2\sqrt{21}\\ =7+4\sqrt{21}\)
21 tháng 4 2021
uses crt;
var a:array[1..100]of integer;
i,n,t:integer;
begin
clrscr;
write('Nhap n='); readln(n);
for i:=1 to n do
begin
write('A[',i,']='); readln(a[i]);
end;
t:=0;
for i:=1 to n do
t:=t+a[i];
writeln(t);
readln;
end.
a) Ta có: \(\dfrac{7x+4}{5}-x=\dfrac{3x-5}{2}\)
\(\Leftrightarrow\dfrac{2\left(7x+4\right)}{10}-\dfrac{10x}{10}=\dfrac{5\left(3x-5\right)}{10}\)
Suy ra: \(14x+8-10x=15x-25\)
\(\Leftrightarrow4x+8-15x+25=0\)
\(\Leftrightarrow-11x+33=0\)
\(\Leftrightarrow-11x=-33\)
hay x=3
Vậy: S={3}
b) ĐKXĐ: \(x\notin\left\{-2;-3\right\}\)
Ta có: \(\dfrac{x-2}{x+2}-\dfrac{x-3}{x+3}+\dfrac{x^2}{\left(x+2\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x+2\right)}{\left(x+3\right)\left(x+2\right)}+\dfrac{x^2}{\left(x+2\right)\left(x+3\right)}=0\)
Suy ra: \(x^2+3x-2x-6-\left(x^2+2x-3x-6\right)+x^2=0\)
\(\Leftrightarrow2x^2+x-6-x^2+x+6=0\)
\(\Leftrightarrow x^2+2x=0\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
Vậy: S={0}