em cần trước 10h ạ
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Những câu hỏi liên quan
NN
17 tháng 1 2022
1 . i wish my mother was here to help me with my homework
2. lan said her brother often travelled abroad with her father.
3 the teacher asked mai if she bought a new calculator
4 it was very generous of john to give them 100
5 if you don't ride a bike fast, you will go to school late
6 it is very important for athletes to be in good health
7 has just made
8 were
9 was having
10 meeting
BT
giúp tui với ạ trước 10h nha tui đang cần gấp
6
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
Bài trên:
\(16x^3y+0,25yz^3=\dfrac{1}{4}y\left(64x^3+z^3\right)=\dfrac{1}{4}y\left[\left(4x\right)^3+z^3\right]\\ =\dfrac{1}{4}y\left[\left(4x+z\right)\left(16x^2-4xz+z^2\right)\right]\\ ----\\ x^4-4x^3+4x^2=x^2\left(x^2-4x+4\right)=x^2\left(x-2\right)^2\\ -----\\ a^3+a^2b-ab^2-b^3=\left(a^3-b^3\right)+\left(a^2b-ab^2\right)\\ =\left(a-b\right)\left(a^2+ab+b^2\right)+ab\left(a-b\right)=\left(a-b\right)\left(a^2+2ab+b^2\right)=\left(a-b\right)\left(a+b\right)^2\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
Bài trên
\(x^3+x^2-4x-4\\ =x^2\left(x+1\right)-4\left(x+1\right)\\ =\left(x^2-4\right)\left(x+1\right)\\ =\left(x-2\right)\left(x+2\right)\left(x+1\right)\\ ---\\ x^3-x^2-x+1\\ =x^2\left(x-1\right)-\left(x-1\right)\\ =\left(x^2-1\right)\left(x-1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x-1\right)=\left(x-1\right)^2\left(x+1\right)\\ ---\\ x^4+x^3+x^2-1\\ =x^3\left(x+1\right)+\left(x-1\right)\left(x+1\right)\\ =\left(x^3+x-1\right)\left(x+1\right)\\ ---\\ x^2y^2+1-x^2-y^2\\ =x^2.\left(y^2-1\right)-\left(y^2-1\right)\\ =\left(y^2-1\right)\left(x^2-1\right)\\ =\left(y-1\right)\left(y+1\right)\left(x-1\right)\left(x+1\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2021
1.
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow x-\dfrac{\pi}{4}=k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
2.
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2021
3.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\dfrac{5}{8}\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^22x=\dfrac{5}{8}\)
\(\Leftrightarrow1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{5}{8}\)
\(\Leftrightarrow\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{5}{8}\)
\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{2\pi}{3}+k2\pi\\4x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\end{matrix}\right.\)
10 tháng 5 2021
Câu 2:
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
if (4<a[i]) and (a[i]<15) then t:=t+a[i];
writeln(t);
readln;
end.
LT
1
H9
HT.Phong (9A5)
CTVHS
4 tháng 10 2023
\(A=\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\) (ĐK: \(x\ge1,x\ne2,x\ne3\))
\(A=\left[\dfrac{\sqrt{x}+\sqrt{x-1}}{\left(\sqrt{x}-\sqrt{x-1}\right)\left(\sqrt{x}+\sqrt{x-1}\right)}-\dfrac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}\right]\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)
\(A=\left[\dfrac{\sqrt{x}+\sqrt{x-1}}{x-x+1}-\dfrac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-1-2}\right]\left[\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right]\)
\(A=\left(\sqrt{x}+\sqrt{x-1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(A=\left(\sqrt{x}-\sqrt{2}\right)\cdot-\dfrac{1}{\sqrt{x}}\)
\(A=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)
b) Ta có: \(x=3-2\sqrt{2}=\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2=\left(\sqrt{2}-1\right)^2\)
Thay vào A ta có:
\(A=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}=\dfrac{\sqrt{2}-\sqrt{2}+1}{\sqrt{2}-1}=\dfrac{1}{\sqrt{2}-1}=\sqrt{2}+1\)
8 tháng 9 2021
1. - Nồi nhôm,khung cửa nhôm, thau nhôm
- Cửa sắt, kéo bằng thép, bồn rửa chén bằng sắt.
- Thau nhựa, ống nước, ghế nhựa.
2. Kẹo hóa học, đường hóa học, thuốc trừ sâu.
3. Thuốc giảm đau, thuốc gây tê, thuốc gây mê, xi măng,...
4. Hộp bút, thước dẻo, bảng đen,...
T
4 tháng 10 2023
\(a,A=\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\)
\(=\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{9+\dfrac{9}{7}-\dfrac{9}{11}+\dfrac{9}{1001}-\dfrac{9}{13}}\)
\(=\dfrac{3\cdot\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}{9\cdot\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}\)
\(=\dfrac{3}{9}\)
\(=\dfrac{1}{3}\)
\(---\)
\(b,B=\dfrac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{7\cdot2^{29}\cdot3^{18}-5\cdot2^{28}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{18}\cdot3^{18}\cdot2^{12}-2\cdot2^{28}\cdot3^{14}\cdot3^4}{2^{28}\cdot3^{18}\cdot\left(7\cdot2-5\right)}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}\cdot\left(14-5\right)}\)
\(=\dfrac{2^{29}\cdot3^{18}\cdot\left(5\cdot2-1\right)}{2^{28}\cdot3^{18}\cdot9}\)
\(=\dfrac{2\cdot\left(10-1\right)}{9}\)
\(=\dfrac{2\cdot9}{9}\)
\(=2\)
\(---\)
\(c,C=\dfrac{5\cdot2^{30}\cdot3^{18}-4\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{2^{28}\cdot3^{18}\cdot\left(5\cdot3-7\cdot2\right)}\)
\(=\dfrac{2^{29}\cdot3^{18}\cdot\left(5\cdot2-3^2\right)}{2^{28}\cdot3^{18}\cdot\left(15-14\right)}\)
\(=\dfrac{2\cdot\left(10-9\right)}{1}\)
\(=2\)
\(---\)
\(d,D=\dfrac{15^{15}\cdot7^{16}}{6\cdot3^{14}\cdot35^{15}-15^8\cdot35^7\cdot7\cdot21^7}\)
\(=\dfrac{\left(3\cdot5\right)^{15}\cdot7^{16}}{2\cdot3\cdot3^{14}\cdot\left(5\cdot7\right)^{15}-\left(3\cdot5\right)^8\cdot\left(5\cdot7\right)^7\cdot7\cdot\left(3\cdot7\right)^7}\)
\(=\dfrac{3^{15}\cdot5^{15}\cdot7^{16}}{2\cdot3^{15}\cdot5^{15}\cdot7^{15}-3^8\cdot5^8\cdot5^7\cdot7^7\cdot7\cdot3^7\cdot7^7}\)
\(=\dfrac{3^{15}\cdot5^{15}\cdot7^{16}}{2\cdot3^{15}\cdot5^{15}\cdot7^{15}-3^{15}\cdot5^{15}\cdot7^{15}}\)
\(=\dfrac{3^{15}\cdot5^{15}\cdot7^{16}}{3^{15}\cdot5^{15}\cdot7^{15}\cdot\left(2-1\right)}\)
\(=\dfrac{7}{1}\)
\(=7\)
#\(Toru\)