GPT:
1) \(\frac{36}{x-12}\)+\(\frac{36}{x}\) =\(\frac{9}{2}\)
2)\(\frac{54}{x+6}+\frac{54}{x-6}=\frac{27}{4}\)
(\(\frac{x+6}{x-6}\))(\(\frac{x+4}{x-4}\))2 +(\(\left(\frac{x-6}{x+6}\right)\left(\frac{x+9}{x-9}\right)^2\)=\(2.\frac{x^2+36}{x^2-36}\)
Cho \(2x-3y=4\)
Tính giá trị của biểu thức A = \(-\frac{8}{3}x^3+\frac{36}{5}x^2y-\frac{54}{5}xy^2+\frac{27}{5}y^3\)
Đề bài lạ thế!
\(A=-\frac{8}{5}x^3+\frac{36}{5}x^2y-\frac{54}{5}xy^2+\frac{27}{5}y^3\)
\(=-\frac{1}{5}\left(8x^3-36x^2y+54xy^2-27y^3\right)\)
=\(-\frac{1}{5}\left(\left(2x\right)^3-3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2-\left(3y\right)^3\right)\)
\(=-\frac{1}{5}\left(2x-3y\right)^3=-\frac{1}{5}.4^3=-\frac{64}{5}\)
Tìm x
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{9}{10}.....\frac{36}{62}.\frac{31}{64}=\frac{1}{2^x}\)
1/4 . 2/6 . 3/8 . ... .30/62 .31/64 = 2^x
(1/2 . 1/2).(2/3 . 1/2).(3/4 . 1/2). ... .(30/31 . 1/2).(31/32 . 1/2) = 2^x
(1/2.1/2. ... .1/2).(1/2 . 2/3 . 3/4. ... .30/31 . 31/32) = 2^x
(31 số 1/2)
(1/2)^31. = 2^x
=> 0=x+36
x=0-36
x=-36
Vậy x=-36
Theo mk nghĩ,mk làm đúng nha .Tk cho mk
Để mk sửa phần này một chút
\((\frac{1}{2})^{31}\cdot\frac{1\cdot2\cdot3.....30\cdot31}{2\cdot3\cdot4.....31\cdot32}=2^x\)
\(\frac{1^{31}}{2^{31}}\cdot\frac{1}{32}=2^x\)
\(\frac{1}{2^{31}}\cdot\frac{1}{2^5}=2^x\)
\(\frac{1}{2^{36}}=2^x\)
\(1=2^x\cdot2^{36}\)
\(2^0=2^x+36\)
Rồi bn tự suy luận nha
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.........................\frac{31}{64}=\frac{1}{2^x}\)
Có 31 thừa số
\(\Leftrightarrow\left(\frac{1}{2}.\frac{1}{2}\right).\left(\frac{1}{2}.\frac{2}{3}\right)...................\left(\frac{1}{2}.\frac{31}{32}\right)=\frac{1}{2^x}\)
\(\Leftrightarrow\left(\frac{1}{2}.\frac{1}{2}...............\frac{1}{2}\right).\left(\frac{1}{2}.\frac{2}{3}...............\frac{31}{32}\right)=\frac{1}{2^x}\)
Có 31 thừa số Có 31 thừa số
\(\Leftrightarrow\frac{1}{2^{31}}.\frac{1.2....................31}{2.3..............32}=\frac{1}{2^x}\)
\(\Leftrightarrow\frac{1}{2^{31}}.\frac{1}{32}=\frac{1}{2^x}\)
\(\Leftrightarrow\frac{1}{2^{31}}.\frac{1}{2^5}=\frac{1}{2^x}\)
\(\Leftrightarrow\frac{1}{2^{36}}=\frac{1}{2^x}\)
\(\Leftrightarrow x=36\)
Vậy x=36
Chúc bạn học tốt
a) \(\left|2x\frac{1}{3}\right|+\frac{5}{6}=1\)
b)\(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
c) \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
d) \(\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=-\frac{53}{10}\)
e) \(\frac{7}{9}:\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
f) \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
g)\(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
h)\(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)
i)\(\frac{6}{2}=\frac{-5+x}{15}\)
k)\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
Câu a \(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)
g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
i) \(\frac{6}{2}=\frac{-5+x}{15}\)
\(\Leftrightarrow3=\frac{x-5}{15}\)
\(\Leftrightarrow x-5=15.3\)
\(\Leftrightarrow x-5=45\)
\(\Leftrightarrow x=45+5\)
\(\Leftrightarrow x=50\)
\(a,\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(b,\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(c,\left(-3\right)^{x+5}=\frac{1}{81}\)
\(d,\left(\frac{1}{9}^x\right)=\left(\frac{1}{27}\right)^6\)
\(e,\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(f,5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(r,4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(h,\left(\frac{1}{2}-\frac{1}{3}\right).6x+6^{x+2}=6^{10}+6^7\)
nhờ mấy bn giúp mk tối mình nộp rồi
a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)
\(\Rightarrow x+3=-3\)
\(\Rightarrow x=-6\)
b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)
\(\Rightarrow2x+2=-2\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)
\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)
\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)
\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
f)\(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
\(5^{x+3}\cdot5-3\cdot5^{x+3}=2\cdot5^{11}\)
\(5^{x+3}\left(5-3\right)=2\cdot5^{11}\)
\(5^{x+3}\cdot2=2\cdot5^{11}\)
\(\Rightarrow5^{x+3}=5^{11}\)
\(\Rightarrow x+3=11\)
\(\Rightarrow x=8\)
r)\(4\cdot3^{x-1}+2\cdot3^{x+2}=4\cdot3^6+2\cdot3^9\)
\(4\cdot3^x:3+2\cdot3^x\cdot9=4.3^7:3+2\cdot3^7\cdot9\)
\(3^x\left(4:3+2\cdot9\right)=3^7\left(4:3+2\cdot9\right)\)
\(\Rightarrow3^x=3^7\)
\(\Rightarrow x=7\)
a)\(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
b)\(\frac{x+6}{2015}+\frac{x+5}{2016}+\frac{x+4}{2017}=\frac{x+3}{2018}+\frac{x+2}{2019}+\frac{x+1}{2010}\)
c)\(\frac{x+6}{2016}+\frac{x+7}{2017}+\frac{x+8}{2018}=\frac{x+9}{2019}+\frac{x+10}{2020}+1\)
d)\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
ai xong nhanh nhất và đúng em xin gửi 2 SP ạ
a) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=0-1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1.\)
Mình chỉ làm câu a) thôi nhé.
Chúc bạn học tốt!
Quy đồng mẫu thức các phần thức sau:
\(\)\(a)\frac{1}{{4{\rm{x}}{y^2}}}\)và \(\frac{5}{{6{{\rm{x}}^2}y}}\);
\(b)\frac{9}{{4{{\rm{x}}^2} - 36}}\)và \(\frac{1}{{{x^2} + 6{\rm{x}} + 9}}\).
\(\)\(a)\frac{1}{{4{\rm{x}}{y^2}}}\)và \(\frac{5}{{6{{\rm{x}}^2}y}}\)
Ta có: MTC là : \(12{{\rm{x}}^2}{y^2}\).
Nhân tử phụ của phân thức \(\frac{1}{{4{\rm{x}}{y^2}}}\)là 3x
Nhân tử phụ của phân thức \(\frac{5}{{6{{\rm{x}}^2}y}}\)là 2y
Khi đó: \(\frac{1}{{4{\rm{x}}{y^2}}} = \frac{{1.3{\rm{x}}}}{{4{\rm{x}}{y^2}.3{\rm{x}}}} = \frac{{3{\rm{x}}}}{{12{{\rm{x}}^2}{y^2}}}\)
\(\frac{5}{{6{{\rm{x}}^2}y}} = \frac{{5.2y}}{{6{{\rm{x}}^2}y.2y}} = \frac{{10y}}{{12{{\rm{x}}^2}{y^2}}}\)
\(b)\frac{9}{{4{{\rm{x}}^2} - 36}}\)và \(\frac{1}{{{x^2} + 6{\rm{x}} + 9}}\).
Ta có: \(\begin{array}{l}4{{\rm{x}}^2} - 36 = 4({x^2} - 9) = 4(x - 3)(x + 3)\\{x^2} + 6{\rm{x}} + 9 = {(x + 3)^2}\end{array}\)
MTC là: \(4(x - 3){(x + 3)^2}\)
Nhân tử phụ của phân thức \(\frac{9}{{4{{\rm{x}}^2} - 36}}\)là: x + 3
Nhân tử phụ của phân thức \(\frac{1}{{{x^2} + 6{\rm{x}} + 9}}\)là 4(x – 3)
Khi đó: \(\begin{array}{l}\frac{9}{{4{{\rm{x}}^2} - 36}} = \frac{9}{{4({x^2} - 9)}} = \frac{9}{{4(x - 3)(x + 3)}} = \frac{{9(x + 3)}}{{4(x - 3){{(x + 3)}^2}}}\\\frac{1}{{{x^2} + 6{\rm{x}} + 9}} = \frac{1}{{{{(x + 3)}^2}}} = \frac{{4(x - 3)}}{{4(x - 3){{(x + 3)}^2}}}\end{array}\)
Giải phương trình
a) \(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
b) \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)
a) \(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
\(\Leftrightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)
\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)
\(\Leftrightarrow\) \(\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)
Vì \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)
\(\Rightarrow x+95=0\)
\(\Leftrightarrow x=-95\)
Vậy phương trình có một nghiệm x = -95
b) \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)
\(\Leftrightarrow\left(\frac{x-1}{59}-1\right)+\left(\frac{x-2}{58}-1\right)+\left(\frac{x-3}{57}-1\right)=\left(\frac{x-4}{56}-1\right)+\left(\frac{x-5}{55}-1\right)+\left(\frac{x-6}{54}-1\right)\)
\(\Leftrightarrow\frac{x-60}{59}+\frac{x-60}{58}+\frac{x-60}{57}-\frac{x-60}{56}-\frac{x-60}{55}-\frac{x-60}{54}=0\)
\(\Leftrightarrow\left(x-60\right)\left(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\right)=0\)
Vì \(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\ne0\)
\(\Rightarrow x-60=0\)
\(\Leftrightarrow x=60\)
Vậy phương trình có một nghiệm x = 60
a) \(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
\(\Rightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)
\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)
\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)
\(\Rightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)
Mà \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)
\(\Rightarrow x+95=0\)
\(\Rightarrow x=-95\)
Vậy x = -95
b) \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)
\(\Rightarrow\left(\frac{x-1}{59}-1\right)+\left(\frac{x-2}{58}-1\right)+\left(\frac{x-3}{57}-1\right)=\left(\frac{x-4}{56}-1\right)+\left(\frac{x-5}{55}-1\right)+\left(\frac{x-6}{54}-1\right)\)
\(\Rightarrow\frac{x-60}{59}+\frac{x-60}{58}+\frac{x-60}{57}-\frac{x-60}{56}-\frac{x-5}{55}-\frac{x-6}{54}=0\)
\(\Rightarrow\left(x-60\right)\left(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\right)=0\)
Mà \(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\ne0\)
\(\Rightarrow x-60=0\)
\(\Rightarrow x=60\)
Vậy x = 60