a} \(\frac{2}{3}.3^{x+1}-7.3^x=-405\)
b} \(\frac{3}{1-2x}=\frac{-5}{3x-2}\)
c} |2x-1| = |2x+3|
giúp mik vs mai mik kiểm tra rùi
a) $\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x2+x}$
b) (x+2).(5-3x)=0
c)$\frac{5(1-2x)}{3}$ + $\frac{x}{2}$ = $\frac{3(x-5)}{4}$ - 2
d)$(x+2)^{2}$ - (x-1).(x+3) = (2x-4).(x+4)-3
e)$(2x-3)^{2}$ = (2x-3).(x+1)
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
Bài 1: Giải các phương trình sau:
a) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
b) \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
Bài 2: Giải các phương trình sau:
a) \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
b) \(\frac{3x-1-\frac{x-1}{2}}{3}-\frac{2x+\frac{1-2x}{3}}{2}=\frac{\frac{3x-1}{2}-6}{5}\)
\(1a,\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(x=-\frac{1}{12}\)
Vậy ................
\(b,\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow\frac{5\left(7x-1\right)}{30}+\frac{30.2x}{30}=\frac{6\left(16-x\right)}{30}\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
Vậy ....................
Bài 1:
c) \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
\(\Leftrightarrow\frac{8.\left(x-2\right)^2}{8.3}-\frac{3.\left(2x-3\right).\left(2x+3\right)}{3.8}+\frac{4.\left(x-4\right)^2}{4.6}=0\)
\(\Leftrightarrow\frac{8.\left(x^2-4x+4\right)}{24}-\frac{3.\left(4x^2-9\right)}{24}+\frac{4.\left(x^2-8x+16\right)}{24}=0\)
\(\Rightarrow8.\left(x^2-4x+4\right)-3.\left(4x^2-9\right)+4.\left(x^2-8x+16\right)=0\)
\(\Leftrightarrow8x^2-32x+32-\left(12x^2-27\right)+4x^2-32x+64=0\)
\(\Leftrightarrow8x^2-32x+32-12x^2+27+4x^2-32x+64=0\)
\(\Leftrightarrow123-64x=0\)
\(\Leftrightarrow64x=123-0\)
\(\Leftrightarrow64x=123\)
\(\Leftrightarrow x=123:64\)
\(\Leftrightarrow x=\frac{123}{64}.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\frac{123}{64}\right\}.\)
Chúc bạn học tốt!
Bài 1:
a) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2-2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x+3=0\)
\(\Leftrightarrow x=12\)
Vậy phương trình có nghiệm là x = 12
Giải các phương trình sau:
a) \(\frac{2x-3}{x^2-9}-\frac{2x}{3-x}=\frac{-5}{x+3}\)
b) \(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{x^2+2x-3}=1\)
a/ ĐK: \(x\ne\pm3\)
\(\frac{2x-3}{x^2-9}+\frac{2x\left(x+3\right)}{x^2-9}+\frac{5\left(x-3\right)}{x^2-9}=0\)
\(\Leftrightarrow2x-3+2x^2+6x+5x-15=0\)
\(\Leftrightarrow2x^2+13x-18=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{-13+\sqrt{313}}{4}\\x=\frac{-13-\sqrt{313}}{4}\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ne1;-3\)
\(\frac{\left(3x-1\right)\left(x+3\right)}{x^2+2x-3}-\frac{\left(2x+5\right)\left(x-1\right)}{x^2+2x-3}+\frac{4}{x^2+2x-3}-\frac{x^2+2x-3}{x^2+2x-3}=0\)
\(\Leftrightarrow3x^2+8x-3-\left(2x^2+3x-5\right)+4-\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow3x+9=0\)
\(\Rightarrow x=-3\) (ko thỏa mãn)
Vậy pt vô nghiệm
Bµi 5: Gi¶i PT sau.
\(a,\frac{5x-2}{2-2x}+\frac{2x-1}{2}+\frac{x^2+x-3}{1-x}=1\)
b,\(\frac{6x-1}{2-x}+\frac{9x+4}{x+2}=\frac{3x^2-2x+1}{x^2-4}\)
\(c,\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
d) (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 = 0
e) x4 + 2x3 + 4x2 + 2x + 1 = 0
\(f,\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{x^2+2x-3}=1\)
a) \(\frac{5x-2}{2-2x}+\frac{2x-1}{2}+\frac{x^2+x-3}{1-x}=1\)
ĐK: x≠1
<=>\(\frac{5x-2}{2\left(1-x\right)}+\frac{2x-1}{2}\frac{x^2+x-3}{1-x}=1\)
<=>\(\frac{5x-2+\left(1-x\right).\left(2x-1\right)+2\left(x^2+x-3\right)}{2\left(1-x\right)}=1\)
<=>\(\frac{5x-2+2x-1-2x^2+x+2x^2+2x-6}{2\left(1-x\right)}=1\)
<=>\(\frac{10x-9}{2\left(1-x\right)}=1\)
<=> 10x-9=2(1-x)
<=>10x-9=2-2x
<=> 10x+2x= 2+9
<=> 12x=11
<=> x= \(\frac{11}{12}\left(tm\right)\)
b) \(\frac{6x-1}{2-x}+\frac{9x+4}{x+2}=\frac{3x^2-2x+1}{x^2-4}\)
ĐK: x≠2, x≠-2
<=>\(\frac{6x-1}{-\left(x-2\right)}+\frac{9x+4}{x+2}-\frac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}=0\)
<=> -(x+2).(6x-1)+(x-2).(9x+4)-(3x2-2x+1)=0
<=> -(6x2-x+12x-2)+9x2+4x-18x-8-3x2+2x-1 = 0
<=> -6x2-11x+2+9x2+4x-18x-8-3x2+2x-1=0
<=> -23x-7=0
<=> -23x=7
<=> x= \(\frac{-7}{23}\left(tm\right)\)
tham khảo câu d trong
https://hoc24.vn/hoi-dap/question/919967.html
c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)
⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
⇒x2+x+1+2x2-5=4x-4
⇔3x2-3x=0
⇔3x(x-1)=0
⇔x=0 (TMĐK) hoặc x=1 (loại)
Vậy tập nghiệm của phương trình đã cho là:S={0}
Bài 2: Giải phương trình:
a) \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
b) \(\frac{3x-1-\frac{x-1}{2}}{3}-\frac{2x+\frac{1-2x}{3}}{2}=\frac{\frac{3x-1}{2}-6}{5}\)
Tìm x
a)\(\left(x-2,5\right)^2\) = \(\frac{4}{9}\)
b)\(\left(2x+\frac{1}{3}\right)^3=\frac{-8}{27}\)
c) \(\frac{2}{3}.3^{x+1}-7.3^x=-405\)
d) \(\left(\frac{3}{5}-\frac{2}{3}x\right)^2=\frac{-64}{125}\)
Giúp mình
a)\(\left(x-2,5\right)^2=\frac{4}{9}\\ \left(x-\frac{5}{2}\right)^2=\left(\pm\frac{2}{3}\right)^2\\\Leftrightarrow\left\{{}\begin{matrix}x-\frac{5}{2}=\frac{2}{3}\\x-\frac{5}{2}=\frac{-2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{19}{6}\\x=\frac{11}{6}\end{matrix}\right. \)
vậy....
b)\(\left(2x+\frac{1}{3}\right)^3=\frac{-8}{27}\\ \left(2x+\frac{1}{3}\right)^3=\left(\frac{-2}{3}\right)^3\\ 2x+\frac{1}{3}=\frac{-2}{3}\\ x=\frac{-1}{2}\)
vậy...
Giải các phương trình sau:
a)\(\left|x^2-3x-5\right|+2\left|2x-1\right|=x^2-4\)
b)\(\frac{4}{2x+1}+\frac{3}{2x+2}=\frac{2}{2x+3}+\frac{1}{2x+4}\)
c)\(\frac{2x-5}{2x^2+3x-5}+\frac{3x+1}{1-x}=\frac{x+20}{4x+10}\)
d)\(\frac{1}{x^2+5x+4}+\frac{1}{x^2+11x+28}+\frac{1}{x^2+17x+70}=\frac{3}{4x-2}\)
Giải các phương trình sau
a,\(\frac{x+5}{3}-\frac{x-3}{5}=\frac{5}{x-3}-\frac{3}{x+5}\)
b, \(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{x^2+2x+3}\)
c,\(\frac{2x}{3x^2-x+2}-\frac{7x}{3x^2+5x+2}=1\)
Các bạn giúp mk nha
a) \(\dfrac{x+5}{3}-\dfrac{x-3}{5}=\dfrac{5}{x-3}-\dfrac{3}{x+5}\)
\(\Rightarrow\dfrac{5\left(x+5\right)}{15}-\dfrac{3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)}{\left(x-3\right)\left(x+5\right)}-\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)
\(\Rightarrow\dfrac{5\left(x+5\right)-3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)
* Với \(5\left(x+5\right)-3\left(x-3\right)=0\),
Ta có được đẳng thức đúng
=> 5x + 25 - 3x + 9 = 0
=> 2x + 34 = 0
=> 2x = -34
=> x = -17
* Với 5( x+5 ) - 3 (x-3 ) \(\ne\)0, ta có
\(\dfrac{5\left(x+5\right)-3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)
\(\Rightarrow\dfrac{1}{15}=\dfrac{1}{\left(x-3\right)\left(x+5\right)}\)
\(\Rightarrow\left(x-3\right)\left(x+5\right)=15\)
\(\Rightarrow x^2+5x-3x-15-15=0\)
\(\Rightarrow x^2+2x-30=0\)
=> \(\left(x+1-\sqrt{31}\right)\left(x+1+\sqrt{31}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1+\sqrt{31}\\x=-1-\sqrt{31}\end{matrix}\right.\)
\(a)\dfrac{x+5}{3}-\dfrac{x-3}{5}=\dfrac{5}{x-3}-\dfrac{3}{x+5}\)(ĐKXĐ: \(x\ne3,x\ne-5\))
\(\Leftrightarrow\dfrac{x+5}{3}-\dfrac{x-3}{5}-\dfrac{5}{x-3}+\dfrac{3}{x+5}=0\\ \Leftrightarrow\dfrac{5\left(x-3\right)\left(x+5\right)^2-3\left(x-3\right)^2\left(x+5\right)-75\left(x+5\right)+45\left(x-3\right)}{15\left(x-3\right)\left(x+5\right)}=0\\ \Leftrightarrow\dfrac{2x^3+38x^2+8x-1020}{15\left(x-3\right)\left(x+5\right)}=0\\ \Leftrightarrow2x^3+38x^2+8x-1020=0\\ \Leftrightarrow\left(x+17\right)\left(x^2+2x-30\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+17=0\\x^2+2x-30=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-17\left(TM\right)\\x=-1+\sqrt{31}\left(TM\right)\\x=-1-\sqrt{31}\left(TM\right)\end{matrix}\right.\)
Vậy....
\(c)\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}=1\)
(ĐKXĐ: \(x\ne-1,x\ne-\dfrac{2}{3}\))
\(\Leftrightarrow\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}-1=0\\ \Leftrightarrow\dfrac{2x\left(3x^2+5x+2\right)-7x\left(3x^2-x+2\right)-\left(3x^2-x+2\right)\left(3x^2+5x+2\right)}{\left(3x^2-x+2\right)\left(3x^2+5x+2\right)}=0\\ \Leftrightarrow\dfrac{-27x^3+10x^2-18x-9x^4-4}{\left(3x^2-x+2\right)\left(3x^2+5x+2\right)}=0\\ \Leftrightarrow-9x^4-27x^3+10x^2-18x-4=0\)
\(\Leftrightarrow-\left(3x^2-2x+2\right)\left(3x^2+11x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-\left(3x^2-2x+2\right)=0\\3x^2+11x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-\left(3x^2-2x+2\right)=0\left(VN\right)\\x=\dfrac{-11-\sqrt{97}}{6}\left(TM\right)\\x=\dfrac{-11+\sqrt{97}}{6}\left(TM\right)\end{matrix}\right.\)
Vậy....
Giải các phương trình sau:
a) \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
b) \(\frac{3x-1-\frac{x-1}{2}}{3}-\frac{2x+\frac{1-2x}{3}}{2}=\frac{\frac{3x-1}{2}}{5}\)
khó quá mk mới học lớp 6 nên k giải đc thông cảm cho mk nha