(x+\(\dfrac{3}{2}\))2=\(\dfrac{9}{49}\)
Những câu hỏi liên quan
1) giải pt :
a) \(\dfrac{7x+10}{x+1}\left(x^2-x-2\right)-\dfrac{7x+10}{x+1}\left(2x^2-3x-5\right)=0\)
b) \(\dfrac{13}{2x^2+x-21}+\dfrac{1}{2x+7}+\dfrac{6}{9-x^2}=0\)
c) \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
d) \(\dfrac{1+\dfrac{x}{x+3}}{1-\dfrac{x}{x+3}}=3\)
a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
=>x=3 hoặc x=-10/7
b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)
\(\Leftrightarrow x^2-12x-51+13x+39=0\)
\(\Leftrightarrow x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=-4
Đúng 0
Bình luận (0)
Bài 1 tìm x
a) (5x1)\(^2\) = \(\dfrac{36}{49}\)
b) ( x- \(\dfrac{2}{9}\) )\(^3\) = ( \(\dfrac{2}{3}\) )\(^6\)
Giúp mình với .
Tìm x,y,z biết:
a) \(\dfrac{x}{5}=\dfrac{y}{2}\) và \(x-y=9\)
b) \(\dfrac{x-3}{12}=\dfrac{-3}{3-x}\)
c) \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{4}\) và \(x-y-z=-49\)
a: Áp dụng tính chất của DTSBN, ta được:
x/5=y/2=(x-y)/(5-2)=9/3=3
=>x=15; y=6
b: =>(x-3)/12=3/(x-3)
=>(x-3)^2=36
=>(x-9)(x+3)=0
=>x=9 hoặc x=-3
c; x/2=y/3
=>x/10=y/15
y/5=z/4
=>y/15=z/12
=>x/10=y/15=z/12=(x-y-z)/(10-15-12)=-49/-17=49/17
=>x=490/17; y=735/17; z=588/17
Đúng 2
Bình luận (0)
1) giải phương trình :
a) left(2+3right)left(dfrac{3x+8}{2-7x}+1right)left(x-5right)left(dfrac{3x+8}{2-7x}+1right)
b) dfrac{7x+10}{x+1}left(x^2-x-2right)-dfrac{7x+10}{x+1}left(2x^2-3x-5right)0
c) dfrac{2x+5}{x+3}+1dfrac{4}{x^2+2x-3}-dfrac{3x-1}{1-x}
d) dfrac{13}{2x^2+x-21}+dfrac{1}{2x+7}+dfrac{6}{9-x^2}0
i) dfrac{x-49}{50}+dfrac{x-50}{49}dfrac{49}{x-50}+dfrac{50}{x-49}
k) dfrac{1+dfrac{x}{x+3}}{1-dfrac{x}{x+3}}3
Đọc tiếp
1) giải phương trình :
a) \(\left(2+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
b) \(\dfrac{7x+10}{x+1}\left(x^2-x-2\right)-\dfrac{7x+10}{x+1}\left(2x^2-3x-5\right)=0\)
c) \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
d) \(\dfrac{13}{2x^2+x-21}+\dfrac{1}{2x+7}+\dfrac{6}{9-x^2}=0\)
i) \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
k) \(\dfrac{1+\dfrac{x}{x+3}}{1-\dfrac{x}{x+3}}=3\)
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
Đúng 0
Bình luận (0)
Giúp mk với mk cảm ơn ạ
a)(x-\(\dfrac{3}{4}\))\(^2\)=0
b)(x+\(\dfrac{4}{9}\))\(^2\)=\(\dfrac{49}{144}\)
a) Ta có: \(\left(x-\dfrac{3}{4}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{3}{4}=0\)
hay \(x=\dfrac{3}{4}\)
b) Ta có: \(\left(x+\dfrac{4}{9}\right)^2=\dfrac{49}{144}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{9}=\dfrac{7}{12}\\x+\dfrac{4}{9}=-\dfrac{7}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{36}\\x=\dfrac{-37}{36}\end{matrix}\right.\)
Đúng 2
Bình luận (0)
\(a,\dfrac{-8}{5}:\left(1+\dfrac{2}{3}\right)\) \(b,\dfrac{7}{5}x\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)
\(c,\dfrac{1}{3}:\left(\dfrac{2}{9}-\dfrac{7}{8}\right)\) \(d,\left(\dfrac{1}{6}-\dfrac{4}{5}\right):\dfrac{7}{5}\)
Giúp mik nha:>>
A -\(\dfrac{24}{25}\)
B -\(\dfrac{5}{21}\)
C -\(\dfrac{24}{47}\)
D -\(\dfrac{19}{42}\)
tick cho mk
Đúng 2
Bình luận (3)
Tìm x :a)dfrac{49}{81}dfrac{7^x}{9} b)dfrac{-64}{343}(dfrac{-4^x}{7})c)dfrac{9}{144}dfrac{3^x}{12} d)dfrac{-1}{32}(dfrac{-1^x}{2})Giúp với ạ bài khó quá . Em cảm ơn ạ !
Đọc tiếp
Tìm x :
a)\(\dfrac{49}{81}\)=\(\dfrac{7^x}{9}\) b)\(\dfrac{-64}{343}\)=(\(\dfrac{-4^x}{7}\))
c)\(\dfrac{9}{144}\)=\(\dfrac{3^x}{12}\) d)\(\dfrac{-1}{32}\)=(\(\dfrac{-1^x}{2}\))
Giúp với ạ bài khó quá . Em cảm ơn ạ !
a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)
b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)
c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)
d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)
Mong bạn xem lại đề bài.
Đúng 3
Bình luận (1)
tìm x
\([\dfrac{6:\dfrac{3}{5}-1\dfrac{1}{16}.\dfrac{16}{7}}{4\dfrac{1}{5}-\dfrac{10}{11}+5\dfrac{2}{11}}-\dfrac{\left(\dfrac{3}{20}+\dfrac{1}{2}-\dfrac{1}{5}\right).\dfrac{12}{49}}{3\dfrac{1}{3}+\dfrac{2}{9}}].x=2\dfrac{23}{96}\)
Hàng dzề, hàng dzề!!! Mại dzô, mại dzô các cậu ơi...Nhanh tay lên kẻo cháy hàng, híhí........................
Giải các phương trình sau
6) left(x^2-1right)^24x+1
7) left(x^2-9right)^212x+1
8) left(x^2+12x+35right)left(x^2-4x+3right)297
9) left(8x+5right)^2left(4x+3right)left(2x+1right)9
10) dfrac{x^2}{3}+dfrac{48}{x^2}10left(dfrac{x}{3}-dfrac{4}{x}right)
11) dfrac{x^4+4}{x^2-2}5x
12) x^4+95xleft(x^2-3right)
13) dfrac{2x}{2x^2-5x+3}+dfrac{13x}{2x^2+x+3}6
14) x^2+dfrac{x^2}{x^2+2x+...
Đọc tiếp
Hàng dzề, hàng dzề!!! Mại dzô, mại dzô các cậu ơi...Nhanh tay lên kẻo cháy hàng, híhí........................
Giải các phương trình sau
6) \(\left(x^2-1\right)^2=4x+1\)
7) \(\left(x^2-9\right)^2=12x+1\)
8) \(\left(x^2+12x+35\right)\left(x^2-4x+3\right)=297\)
9) \(\left(8x+5\right)^2\left(4x+3\right)\left(2x+1\right)=9\)
10) \(\dfrac{x^2}{3}+\dfrac{48}{x^2}=10\left(\dfrac{x}{3}-\dfrac{4}{x}\right)\)
11) \(\dfrac{x^4+4}{x^2-2}=5x\)
12) \(x^4+9=5x\left(x^2-3\right)\)
13) \(\dfrac{2x}{2x^2-5x+3}+\dfrac{13x}{2x^2+x+3}=6\)
14) \(x^2+\dfrac{x^2}{x^2+2x+1}=3\)
15) \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
Tính:
a) \(\dfrac{3}{14}\) của -49;
b) \(\dfrac{3}{4}\) của \(\dfrac{-18}{25}\);
c) \(1\dfrac{2}{3}\) của \(3\dfrac{2}{9}\);
d) 40% của \(\dfrac{20}{9}\).
\(\dfrac{3}{14}\cdot\left(-49\right)=-\dfrac{21}{2}\)
\(\dfrac{3}{4}\cdot\dfrac{-18}{25}=-\dfrac{27}{50}\)
\(1\dfrac{2}{3}\cdot3\dfrac{2}{9}=\dfrac{29}{9}\cdot\dfrac{5}{3}=\dfrac{145}{27}\)
\(40\%\cdot\dfrac{20}{9}=\dfrac{40}{100}\cdot\dfrac{20}{9}=\dfrac{40}{45}=\dfrac{8}{9}\)
Đúng 1
Bình luận (0)
a, \(-49.\dfrac{3}{14}=-\dfrac{21}{2}\)
b, \(\dfrac{-18}{25}.\dfrac{3}{4}=-\dfrac{27}{50}\)
c, \(3\dfrac{2}{9}.1\dfrac{2}{3}=\dfrac{29}{9}.\dfrac{5}{3}=\dfrac{145}{27}\)
d, \(\dfrac{20}{9}.40\%=\dfrac{20}{9}.\dfrac{40}{100}=\dfrac{8}{9}\)
Đúng 0
Bình luận (0)