C = \(\dfrac{2x-7}{x+1}\) D = \(\dfrac{5x+9}{x+3}\)
Mình đang cần gấp
Giải phương trình sau :
a,\(\dfrac{7-3x}{12}+\dfrac{5x+2}{7}=x+13\)
b,\(\dfrac{3\left(x+3\right)}{4}-\dfrac{1}{2}=\dfrac{5x+9}{7}-\dfrac{7x-9}{4}\)
c,\(\dfrac{2x+1}{3}-\dfrac{5x+2}{7}=x+3\)
d,\(\dfrac{2x-3}{3}-\dfrac{2x+3}{7}=\dfrac{4x+3}{5}-17\)
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
a, msc 12.7=84
Chuyển vế về =0 rồi làm
b,msc 28
c,làm tương tự
a, \(\Rightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow-45x=1019\Leftrightarrow x=-\dfrac{1019}{45}\)
b, \(\Rightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
\(\Leftrightarrow21x+63-14=20x+36-49x+63\)
\(\Leftrightarrow50x=50\Leftrightarrow x=1\)
c, \(\Rightarrow14x+7-15x-6=21x+63\Leftrightarrow-22x=62\Leftrightarrow x=-\dfrac{31}{11}\)
d, \(\Rightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-105.17\)
\(\Leftrightarrow70x-105-30x-45=84x+63-1785\)
\(\Leftrightarrow-44x=-1572\Leftrightarrow x=\dfrac{393}{11}\)
Giải phương trình
a)\(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
b)\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)
c)\(\left(\dfrac{3}{2x+1}+2\right)\left(5x-2\right)=\dfrac{5x-2}{2x+1}\)
d)\(\dfrac{2x-1}{x^3+1}+\dfrac{1}{x+1}=\dfrac{2}{x^2-x+1}\)
GIÚP MÌNH VỚI , ĐANG CẦN GẤP
a.x-\(\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
⇔\(x=\dfrac{7-3x}{4}+\dfrac{5x+2}{6}\)
⇔\(x=\dfrac{21-9x+10x+4}{12}\)
⇔x=\(\dfrac{x+25}{12}\)
⇔12x=x+25
⇔x=\(\dfrac{25}{11}\)
Vậy pt đã cho có n0 là S=\(\left\{\dfrac{25}{11}\right\}\)
b.ĐKXĐ:x≠-2;x≠2
\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)
⇔\(\dfrac{\left(x-2\right)\cdot\left(x-2\right)-3\cdot\left(x+2\right)}{\left(x-2\right)\cdot\left(x+2\right)}\)=\(\dfrac{2x-22}{\left(x-2\right)\cdot\left(x+2\right)}\)
⇔\(\dfrac{x^2-7x-2}{\left(x-2\right)\cdot\left(x+2\right)}=\dfrac{2x-22}{\left(x-2\right)\cdot\left(x+2\right)}\)
⇒\(\left(x^2-7x-2\right)\cdot\left(x-2\right)\cdot\left(x+2\right)=\left(2x-22\right)\cdot\left(x-2\right)\cdot\left(x+2\right)\)
⇔x2-7x-2=2x-22
⇔x2-9x+20=0
⇔(x-4)(x-5)=0
⇔\(\left\{{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Vậy pt đã cho có n0 là S={4;5}
\(\left(\dfrac{3}{2x+1}+\dfrac{4x+2}{2x+1}\right)\cdot\left(5x-2\right)=\dfrac{5x-2}{2x+1}\)
⇔\(\dfrac{\left(4x+5\right)\cdot\left(5x-2\right)}{2x+1}=\dfrac{5x-2}{2x+1}\)
⇒(4x+5)(5x-2)(2x+1)=(5x-2)(2x+1)
⇔4x+5=1
⇔x=-1
Vậy pt đã cho có n0 là S={-1}
Viết phương trình đường thẳng (d) đi qua hai điểm A (-2;1) và M. Biết M nằm trên đường thẳng (d') y= 3-2x và M có hoành độ bằng \(\dfrac{1}{2}\)?
A. y = 2x+5
B. y = -2x-3
C. y = \(\dfrac{2}{5}x\)+\(\dfrac{9}{5}\)
D. y = 5x-3
Mình đang cần rất gấp ạ! Cảm ơn mn nhiều.
có thể giúp mình giải bài này với đc k ạ mình đang cần gấp (xin cảm ơn)
Bài 1:
a,\(3x-7\sqrt{x}+4=0\)
b, \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
c, \(\dfrac{\sqrt{x}-2}{\sqrt{x}-4}=\dfrac{6-\sqrt{x}}{7-\sqrt{x}}\)
d, \(\sqrt{x-3}-\dfrac{5}{3}\sqrt{9x-27}+\dfrac{3}{2}\sqrt{4x-12}=-1\)
Bài 2:
a, \(\sqrt{x^2+6x+9}=3x-6\)
b, \(\sqrt{3x^2}=x+2\)
c, \(\sqrt{x^2-4x+4}-2x+5=0\)
d, \(x^2-2\sqrt{7x}+7=0\)
Bài 3:
a, \(\sqrt{3+x}+\sqrt{6-x}=3\)
b, \(\sqrt{3+x}-\sqrt{2-x}=1\)
Bài 2
b, `\sqrt{3x^2}=x+2` ĐKXĐ : `x>=0`
`=>(\sqrt{3x^2})^2=(x+2)^2`
`=>3x^2=x^2+4x+4`
`=>3x^2-x^2-4x-4=0`
`=>2x^2-4x-4=0`
`=>x^2-2x-2=0`
`=>(x^2-2x+1)-3=0`
`=>(x-1)^2=3`
`=>(x-1)^2=(\pm \sqrt{3})^2`
`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$
Vậy `S={1+\sqrt{3};1-\sqrt{3}}`
Bài 1
a, `3x-7\sqrt{x}+4=0` ĐKXĐ : `x>=0`
`<=>3x-3\sqrt{x}-4\sqrt{x}+4=0`
`<=>3\sqrt{x}(\sqrt{x}-1)-4(\sqrt{x}-1)=0`
`<=>(3\sqrt{x}-4)(\sqrt{x}-1)=0`
TH1 :
`3\sqrt{x}-4=0`
`<=>\sqrt{x}=4/3`
`<=>x=16/9` ( tm )
TH2
`\sqrt{x}-1=0`
`<=>\sqrt{x}=1` (tm)
Vậy `S={16/9;1}`
b, `1/2\sqrt{x-1}-9/2\sqrt{x-1}+3\sqrt{x-1}=-17` ĐKXĐ : `x>=1`
`<=>(1/2-9/2+3)\sqrt{x-1}=-17`
`<=>-\sqrt{x-1}=-17`
`<=>\sqrt{x-1}=17`
`<=>x-1=289`
`<=>x=290` ( tm )
Vậy `S={290}`
Bài 1:
a) Ta có: \(3x-7\sqrt{x}+4=0\)
\(\Leftrightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)
b) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}\cdot\left(-1\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
a, x =\(\dfrac{-2}{7}\) +\(\dfrac{9}{7}\)
b,\(\dfrac{x}{3}\) =\(\dfrac{2}{5}\) +\(\dfrac{-4}{3}\)
Làm giúp mik vs mình đang cần gấp!
a) \(x=\dfrac{-2}{7}+\dfrac{9}{7}=1\)
b) \(\dfrac{x}{3}=\dfrac{2}{5}+\dfrac{-4}{3}\)
\(\dfrac{x}{3}=\dfrac{-14}{15}\)
\(\Rightarrow x=\dfrac{3.-14}{15}=\dfrac{-14}{5}\)
\(x=\dfrac{-2}{7}+\dfrac{9}{7}\)
\(x=1\)
3) tìm x biết
a) \(\sqrt{x+9}=7\)
b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\)
c) \(\sqrt{x^2-6x+9}=2x+1\)
d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\)
lm nhanh giúp mk nhé mk đang cần gấp
Lời giải:
a. ĐKXĐ: $x\geq -9$
PT $\Leftrightarrow x+9=7^2=49$
$\Leftrightarrow x=40$ (tm)
b. ĐKXĐ: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$
$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$
$\Leftrgihtarrow 3\sqrt{2x+3}=15$
$\Leftrightarrow \sqrt{2x+3}=5$
$\Leftrightarrow 2x+3=25$
$\Leftrightarrow x=11$ (tm)
c.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{2}{3}\)
d. ĐKXĐ: $x\geq 1$
PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)
\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)
\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)
\(\Leftrightarrow -1=9\) (vô lý)
Vậy pt vô nghiệm.
a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)
b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)
\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)
\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)
\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)
c) \(\sqrt{x^2-6x+9}=2x+1\)
Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)
\(\Rightarrow\sqrt{\left(x-3\right)^2}=2x+1\Rightarrow\left|x-3\right|=2x+1\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)
d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)
\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)
\(\Rightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}-\sqrt{\left(\sqrt{x-1}+3\right)^2}=9\)
\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)
\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)
Giải các phương trình sau:
a) 2,3 - 2(0,7 + 2) = 3,6 - 1,7x
b) \(\dfrac{5x+7}{4}-\dfrac{3x+5}{8}=\dfrac{4x+9}{5}-\dfrac{x-9}{3}\)
c) \(\dfrac{2x-1}{4}+\dfrac{x-3}{3}=\dfrac{4x-2}{3}-\dfrac{6x+7}{12}\)
d) (x - 1)(x + 2) - x(x + 3) = 8
a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1
=>1,7x=6,7
hay x=67/17
b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)
=>150x+120-45x-75=96x+216-40x+360
=>105x+45=56x+576
=>49x=531
hay x=531/49
3)tìm GTLN của \(A=\dfrac{\sqrt{x-9}}{5x}\)
giúp mình vs mk đang cần gấp
\(A=\dfrac{\sqrt{x-9}}{5x}\left(ĐKx\ge9\right)\)
A'=\(\dfrac{\dfrac{5x}{2\sqrt{x-9}}-5\sqrt{x-9}}{\left(5x^2\right)}\)
\(A'=0\rightarrow5x=10\left(x-9\right)\)
\(\rightarrow x=18\)
\(MaxA=\dfrac{1}{30}\) khi \(x=18\)
\(A=\dfrac{2.3\sqrt{x-9}}{30x}\le\dfrac{3^2+x-9}{30x}=\dfrac{1}{30}\)
\(A_{max}=\dfrac{1}{30}\) khi \(\sqrt{x-9}=3\Leftrightarrow x=18\)
\(a.\dfrac{5x-20}{3}-\dfrac{x\left(1-3x\right)}{3}-\dfrac{5x}{4}\)
b.\(\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
c.\(\dfrac{-1}{2x+3}< 0\)
CỨU MÌNH VỚI MỌI NGƯỜI MÌNH ĐANG CẦN GẤP MAI MINK KTR RỒI. HELP ME!!!!!!!!!!
a. có vấn đề
b.
\(\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow20x^2-12x+15x+5< 20x^2+10x-30\)
\(\Leftrightarrow-22x+5x< -30-5\)
\(\Leftrightarrow-17x< -35\)
\(\Leftrightarrow x>\dfrac{35}{17}\)