\(\left(\frac{1}{2}-\frac{1}{6}\right).3^x+3^{x+1}=3^{16}+3^{13}\\ \frac{1}{3}.3^x+3^{x+1}=3^{16}+3^{13}\\ 3^{x-1}+3^{x+1}=3^{16}+3^{13}\)
Thay \(x-1=16\) có \(x=17\)
Thay \(x+1=13\) có \(12\)
vậy \(x=\left\{17;12\right\}\)
chắc thế, sai kệ nhá
Tìm x biết: \(\frac{4}{\left(x+2\right).\left(x+6\right)}+\frac{7}{\left(x+6\right).\left(x+13\right)}=\frac{2x+1}{\left(x+2\right).\left(x+16\right)}-\frac{3}{\left(x+13\right).\left(x+16\right)}\)
Tìm x biết: \(\frac{4}{\left(x+2\right).\left(x+6\right)}+\frac{7}{\left(x+6\right).\left(x+13\right)}=\frac{2x+1}{\left(x+2\right).\left(x+16\right)}-\frac{3}{\left(x+13\right).\left(x+16\right)}\)
Tìm x: \(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}+\frac{1}{x-28}=\frac{-1}{20}\)
Tìm tập hợp các số nguyên x thỏa mãn :
a, \(3\frac{1}{3}:2\frac{1}{2}-1< x< 7\frac{2}{3}.\frac{3}{7}+\frac{5}{2}\)
b,\(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)
3/a)\(M\left(x\right)=\frac{1}{2}x^2-3x-x^3+3\)
\(=-x^3+\frac{1}{2}x^2-3x+3\)
\(N\left(x\right)=-4x+x^2+\frac{1}{2}x^3+6\)
\(=\frac{1}{2}x^3+x^2-4x^3+6\)
b)Ta có:\(\text{A}\left(x\right)=M\left(x\right)-N\left(x\right)\)
hay \(\text{A}\left(x\right)=\left(-x^3+\frac{1}{2}x^2-3x+3\right)-\left(\frac{1}{2}x^3+x^2-4x+6\right)\)
\(=-x^3+\frac{1}{2}x^2-3x+3-\frac{1}{2}x^3-x^2+4x-6\)
\(=-\frac{3}{2}x^3-\frac{1}{2}x^2+x-3\)
Đặt\(\text{A}\left(x\right)=0\)
\(\Rightarrow-\frac{3}{2}x^3-\frac{1}{2}x^2+x-3=0\)
\(-\frac{3}{2}x^3-\frac{1}{2}x^2=-x+3\)
\(-2\left(x^3-x^2\right)=-x+3\)
\(x^3-x^2+x=3+2=5\)
\(x^2=5\)
\(\Rightarrow x=\sqrt{5}\)
Vậy \(\text{A}\left(x\right)\) có 1 nghiệm là \(\sqrt{5}\)
1)Tính
a)\(\frac{\left(3^3\right)^2\times\left(2^3\right)^5}{\left(2\times3\right)^6\times\left(2^5\right)^3}\)
2)So sánh
\(5^{1000}\)và\(3^{1500}\)
3)Tìm x biết
\(\left(\frac{1}{2}\right)^{x+1}=\frac{1}{16}\)
\(\left(\frac{2}{3}x-1\right).\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)
\(\left(x-1\right)^{^{ }x-2}=\left(x-1\right)^{x+6}\)
Bài 1: Thu gọn
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
c) \(\frac{1}{7}x^2y^3.\left(-\frac{14}{3}xy^2\right)-\frac{1}{2}xy.\left(x^2y^{\text{4}}\right)\)
d) \(\left(3xy\right)^2.\left(-\frac{1}{2}x^3y^2\right)\)
e) \(-\frac{1}{4}xy^2+\frac{2}{5}x^2y+\frac{1}{2}xy^2-x^2y\)
f) \(\frac{1}{2}x^4y.\left(-\frac{2}{3}x^3y^2\right)-\frac{1}{3}x^7y^3\)
g) \(\frac{1}{2}x^2y.\left(-10x^3yz^2\right).\frac{1}{4}x^5y^3z\)
h) \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)
i) \(1\frac{2}{3}x^3y.\left(\frac{-1}{2}xy^2\right)^2-\frac{5}{4}.\frac{8}{15}x^3y.\left(-\frac{1}{2}xy^2\right)^2\)
k) \(-\frac{3}{2}xy^2.\left(\frac{3}{4}x^2y\right)^2-\frac{3}{5}xy.\left(-\frac{1}{3}x^4y^3\right)+\left(-x^2y\right)^2.\left(xy\right)^2\)
n) \(-2\frac{1}{5}xy.\left(-5x\right)^2+\frac{3}{4}y.\frac{2}{3}\left(-x^3\right)-\frac{1}{9}.\left(-x\right)^3.\frac{1}{3}y\)
m) \(\left(-\frac{1}{3}xy^2\right)^2.\left(3x^2y\right)^3.\left(-\frac{5}{2}xy^2z^3\right)^{^2}\)
p) \(-2y.\left|2\right|x^4y^5.\left|-\frac{3}{4}\right|x^3y^2z\)
Bài toán khó nhất thế kỉ đố ai giải được
1!) \(\frac{1}{2}\)X<2 và \(\frac{1}{X}\)>-3
2@) M=\(\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2}}}}\)
3) Tìm X, biết rằng nếu lấy 1 trừ đi số nghịch đảo của 1-X ta lại được số nghịch đảo của 1-X
4) \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+......+\frac{1}{16}\left(1+2+3+...+16\right)\)
(Đề thi học sinh giỏi của Mỹ năm 2022)
