có bạn nào giúp mk dc ko? 11h mk pk lên trường r
tìm x biết \(3\dfrac{1}{2}:|2x-1|=\dfrac{21}{22}\)
Giải các phương trình sau: (TM ĐK)
1) \(\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)
2) \(\dfrac{2x-1}{5-3x}=2\)
3) \(\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)
4) \(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
mng giúp mk bài này vs. Cảm ơn bạn nhiều
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\left(ĐKXĐ:x\ne5\right)\)
\(\Rightarrow3\left(4x-3\right)=29\left(x-5\right)\)
\(\Leftrightarrow12x-9=29x-145\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x+136=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\left(tm\right)\)
Vậy \(S=\left\{8\right\}\)
\(2,\dfrac{2x-1}{5-3x}=2\left(ĐKXĐ:x\ne\dfrac{5}{3}\right)\)
\(\Rightarrow2x-1=2\left(5-3x\right)\)
\(\Leftrightarrow2x-1=10-6x\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x-11=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{11}{8}\right\}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(ĐKXĐ:x\ne1\right)\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2\left(x-1\right)}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2x-2}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{3x-2}{x-1}\)
\(\Rightarrow4x-5=3x-2\)
\(\Leftrightarrow4x-5-3x+2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy \(S=\left\{3\right\}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne-5\right)\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2+15x+25}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{15x+25}{2x\left(x+5\right)}=0\)
\(\Rightarrow15x+25=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{-5}{3}\right\}\)
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-29\left(x-5\right)}{3\left(x-5\right)}=0\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\)
\(2,\dfrac{2x-1}{5-3x}=2\)
\(\Leftrightarrow\dfrac{2x-1-2\left(5-3x\right)}{5-3x}=0\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5-2\left(x-1-x\right)}{x-1}=0\)
\(\Leftrightarrow4x-5-2x+2+2x=0\)
\(\Leftrightarrow4x=3\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
tìm x biết
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}.x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
|x|\(-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\left|2.x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
giúp mk vs nhanh lên mình đang bận
b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)
c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)
Tìm x, biết:
a) \(\left|x-3\right|-2x=1\)
b) \(\dfrac{4^{x+2}+4^{x+1}+4^x}{21}=\dfrac{3^{2x}+3^{2x+1}+3^{2x+3}}{31}\)
Các bn giúp mk nhanh nha. Tối nay mk phải nộp cho thầy r. Mk sẽ tick cho.
a)TH1 x>=3 \(\left|x-3\right|\)=x-3
pttt: x-3-2x=1 suy ra x=-4 <3 -> loại
TH2 x=< 3 pttt 3-x-2x=1 suy ra x =2/3 thỏa mãn
b) VT=\(\dfrac{4^{x+2}+4^{x+1}+4^x}{21}=\dfrac{4^x\left(4^2+4+1\right)}{21}=4^x\)
VP= \(\dfrac{3^{2x}+3^{2x+1}+3^{2x+3}}{31}=\dfrac{9^x\left(1+3+27\right)}{31}=9^x\)
vậy pt đã cho tương đương với 4^x=9^x \(\Leftrightarrow\left(\dfrac{4}{9}\right)\)^x =1 suy ra x =0
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)
\(\dfrac{x+1}{22}=\dfrac{6}{x}\)
\(\dfrac{2x-1}{2}=\dfrac{5}{x}\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\)
\(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\)
Tìm x
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)
`=> (x-3)5 = (2x+1)3`
`=> 5x-15 = 6x+3`
`=> 5x-6x = 15+3`
`=> -x=18`
`=> x=-18`
\(\dfrac{x+1}{22}=\dfrac{6}{x}\)
`=> (x+1)x = 22*6`
`=> (x+1)x = 132`
`=> x^2 + x = 132`
`=> x^2+x-132=0`
`=> (x-11)(x+12)=0`
`=>`\(\left[{}\begin{matrix}x-11=0\\x+12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=11\\x=-12\end{matrix}\right.\)
\(\dfrac{2x-1}{2}=\dfrac{5}{x}\)
`=> (2x-1)x = 2*5`
`=> 2x^2 - x =10`
`=> 2x^2 - x - 10 =0`
`=> 2x^2 + 4x - 5x - 10 =0`
`=> (2x^2 + 4x) - (5x+10)=0`
`=> 2x(x+2) - 5(x+2)=0`
`=> (2x-5)(x+2)=0`
`=>`\(\left[{}\begin{matrix}2x-5=0\\x+2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\)
`=> (2x-1)(2x+1)=21*3`
`=> 4x^2 + 2x - 2x - 1 = 63`
`=> 4x^2 - 1=63`
`=> 4x^2 - 1 - 63=0`
`=> 4x^2 - 64 = 0`
`=> 4(x^2 - 16)=0`
`=> 4(x^2 + 4x - 4x - 16)=0`
`=> 4[(x^2+4x)-(4x+16)]=0`
`=> 4[x(x+4)-4(x+4)]=0`
`=> 4(x-4)(x+4)=0`
`=>`\(\left[{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
\(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\)
`=> (2x+1)(x+1) = 9*5`
`=> (2x+1)(x+1)=45`
`=> 2x^2 + 2x + x + 1 = 45`
`=> 2x^2 + 3x + 1 =45`
`=> 2x^2 + 3x + 1 - 45 =0`
`=> 2x^2+3x-44=0`
`=> 2x^2 + 11x - 8x - 44=0`
`=> (2x^2 +11x) - (8x+44)=0`
`=> x(2x+11) - 4(2x+11)=0`
`=> (x-4)(2x+11)=0`
`=>`\(\left[{}\begin{matrix}x-4=0\\2x+11=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\\ \left(x-3\right)\cdot5=\left(2x+1\right)\cdot3\\ x5-15=6x+3\\ x5-6x=3+15\\ -x=18\\ \Rightarrow x=-18\)
\(\dfrac{x+1}{22}=\dfrac{6}{x}\\ \left(x+1\right)\cdot x=6\cdot22\\ \left(x+1\right)\cdot x=2\cdot3\cdot2\cdot11\\ \left(x+1\right)\cdot x=12\cdot11\\ \Rightarrow x=11\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\\ \left(2x-1\right)\cdot\left(2x+1\right)=21\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot3\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot9\\ \Rightarrow2x+1=9\\ 2x=8\\ x=4\)
Tìm x:
3\(\dfrac{1}{14}\) - 5\(\dfrac{1}{3}\) - \(\dfrac{4}{7}\) + \(\dfrac{11}{21}\) = x -\(\dfrac{1}{2}\)
Làm ra bài giải giúp mk vs
=>x-1/2=3+1/14-5-1/3-4/7+11/21
=>x-1/2=-97/42
=>x-21/42=-97/42
=>x=-97/42+21/42=-76/42=-38/21
giúp mk vs :(( ( tìm x )
\(\dfrac{-4}{x}=\dfrac{x}{-49}\); \(\dfrac{3.5}{x-3}=\dfrac{5}{3}\)
(2x + 1) : 2 = 12 : 3 ; ( 2x - 14 ): 3= 12 :9
giúp mk nhaa vội lém á :((
\(\dfrac{-4}{x}=\dfrac{x}{-49}\\ \Rightarrow x^2=\left(-4\right)\left(-49\right)\\ \Rightarrow x^2=196\\ \Rightarrow x=\pm14\)
\(\dfrac{3.6}{x-3}=\dfrac{5}{3}\\ \Rightarrow5\left(x-3\right)=3.3.6\\ \Rightarrow5\left(x-3\right)=54\\ \Rightarrow x-3=\dfrac{54}{5}\\ \Rightarrow x=\dfrac{54}{5}+3\\ \Rightarrow x=\dfrac{69}{15}\)
\(\left(2x+1\right):2=12:3\\ \left(2x+1\right):2=4\\2x+1=2\\ 2x=1\\ x=\dfrac{1}{2} \)
\(\left(2x-14\right):3=12:9\\ \left(2x-14\right):3=\dfrac{4}{3}\\ 2x-14=4\\ 2x=16\\ x=8\)
Tìm điều kiện có nghĩa:
1) \(\sqrt{x^2+2x-3}\)
2) \(\sqrt{2x^2+5x+3}\)
3) \(\sqrt{\dfrac{4}{x-1}}\)
4) \(\sqrt{\dfrac{-1}{x-3}}\)
5) \(\sqrt{\dfrac{-3}{x+2}}\)
6) \(\sqrt{\dfrac{1}{2a-1}}\)
LÀM CHI TIẾT GIÚP MK NHÉ!
1) ĐKXĐ: \(x^2+2x-3\ge0\Leftrightarrow\left(x+1\right)^2\ge4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1\ge2\\x+1\le-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\)
2) ĐKXĐ: \(2x^2+5x+3\ge0\Leftrightarrow2\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{8}\Leftrightarrow\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{4}\ge\dfrac{1}{4}\\x+\dfrac{5}{4}\le-\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge-1\\x\le-\dfrac{3}{2}\end{matrix}\right.\)
3) ĐKXĐ: \(x-1>0\Leftrightarrow x>1\)
4) ĐKXĐ: \(x-3< 0\Leftrightarrow x< 3\)
5) ĐKXĐ: \(x+2< 0\Leftrightarrow x< -2\)
6) ĐKXĐ: \(2a-1>0\Leftrightarrow a>\dfrac{1}{2}\)
\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\)
Giúp mk với
`1/(x-1)-(3x^2)/(x^3-1)=(2x)/(x^2+x+1)`
ĐK:`x ne 1`
`pt<=>(x^2+x+1)/(x^3-1)-(3x^2)/(x^3-1)=(2x(x-1))/(x^3-1)`
`<=>x^2+x+1-3x^2=2x^2-2x`
`<=>4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>4x(x-1)+x-1=0`
`<=>(x-1)(4x+1)=0`
`x ne 1=>x-1 ne 0`
`<=>4x+1=0`
`<=>x=-1/4`
Vậy `S={-1/4}`
đk: x khác 1
\(\dfrac{1}{x-1}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2x}{x^2+x+1}=0\)
<=> \(\dfrac{x^2+x+1-3x^2-2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
<=> \(x^2+x+1-3x^2-2x^2+2x=0\)
<=> -4x2 + 3x + 1 =0
<=> (1-x)(4x+1) = 0
Mà x khác 1
<=> 4x + 1 = 0
<=> \(x=\dfrac{-1}{4}\)
(x^2+x+1)/(x^3-1)-(3x^2)/(x^3-1)=(2x)*(x-1)/(x^3-1)
=>(x^2+x+1-3x^2-2x^2+2x)=0
=>(-4x^2+3x+1)=0
mua máy casio về bấm giải ptr ra
Tìm giá trị nhỏ nhất của biểu thức sau :
\(A=|2x-\dfrac{1}{3}|-1\dfrac{3}{4}\)
\(B=\dfrac{1}{3}|x-2|+|3-\dfrac{1}{2}y|+4\)
Giúp mk với ! Mk cần gấp lắm !!!
a) Ta có: \(\left|2x-\dfrac{1}{3}\right|\ge0\forall x\)
\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|-\dfrac{7}{4}\ge-\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi \(2x=\dfrac{1}{3}\)
hay \(x=\dfrac{1}{6}\)
Vậy: \(A_{min}=-\dfrac{7}{4}\) khi \(x=\dfrac{1}{6}\)
b) Ta có: \(\dfrac{1}{3}\left|x-2\right|\ge0\forall x\)
\(\left|3-\dfrac{1}{2}y\right|\ge0\forall y\)
Do đó: \(\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|\ge0\forall x,y\)
\(\Leftrightarrow\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|+4\ge4\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\3-\dfrac{1}{2}y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)
Vậy: \(B_{min}=4\) khi x=2 và y=6