tìm x biết
4.(2- |x|)+5. |x|=7
b) |5x-3|= |7-x|
c) |\(\frac{1}{2}x-3\)|=|\(\frac{1}{3}x-2\)|
d)|x-2|-2x=0,25
e)|15+x|+x=-15
giải phương trình:\(\frac{2x}{6x^2-x+3}+\frac{5x}{4x^2+5x+2}+\frac{x}{2x^2+3x+1}=\frac{1}{3}\)
b, \(\frac{1}{x+1}+\frac{2}{x+2}+\frac{1}{x+3}=\frac{1}{x+4}+\frac{2}{x+5}+\frac{1}{x+6}\)
c, \(x^2+\frac{9x^2}{\left(x+3\right)^2}=7\)
d,\(\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}=0\)
e,\(\frac{9x}{x^2-2x+3}=\frac{5x^2+9x+15}{x^2+3x+3}\)
a,ĐKXĐ \(x\ne-1;-\frac{1}{2}\)
Ta thấy x=0 không là nghiệm của PT
Xét \(x\ne0\)
Khi đó PT
<=> \(\frac{2}{6x-1+\frac{3}{x}}+\frac{5}{4x+5+\frac{2}{x}}+\frac{1}{2x+3+\frac{1}{x}}=\frac{1}{3}\)
Đặt \(2x+\frac{1}{x}=a\)
=> \(\frac{2}{3a-1}+\frac{5}{2a+5}+\frac{1}{a+3}=\frac{1}{3}\)
<=> \(3\left(25a^2+75a+10\right)=6a^3+31a^2+34a-15\)
<=> \(6a^3-44a^2-191a-45=0\)
Xin lỗi đến đây tớ ra nghiệm không đẹp
c, \(x^2+\frac{9x^2}{\left(x+3\right)^2}=7\) ĐKXĐ \(x\ne-3\)
<=> \(\left(x-\frac{3x}{x+3}\right)^2+2.\frac{3x^2}{x+3}=7\)
<=> \(\left(\frac{x^2}{x+3}\right)^2+6.\frac{x^2}{x+3}-7=0\)
<=> \(\left(\frac{x^2}{x+3}+7\right)\left(\frac{x^2}{x+3}-1\right)=0\)
<=> \(\orbr{\begin{cases}x^2+7x+21=0\\x^2-x-3=0\end{cases}}\)
\(S=\left\{\frac{1\pm\sqrt{13}}{2}\right\}\)thỏa mãn ĐKXĐ
b,\(\frac{1}{x+1}+\frac{2}{x+2}+\frac{1}{x+3}=\frac{1}{x+4}+\frac{2}{x+5}+\frac{1}{x+6}\)ĐKXĐ \(x\ne-1;-2;-3;-4;-5;-6\)
<=>\(\left(\frac{1}{x+1}-\frac{1}{x+6}\right)+2\left(\frac{1}{x+2}-\frac{1}{x+5}\right)+\left(\frac{1}{x+3}-\frac{1}{x+4}\right)=0\)
<=>\(\frac{5}{x^2+7x+6}+\frac{6}{x^2+7x+10}+\frac{1}{x^2+7x+12}=0\)
Đặt \(x^2+7x+6=a\)
=> \(\frac{5}{a}+\frac{6}{a+4}+\frac{1}{a+6}=0\)
<=> \(12a^2+90a+120=0\)
<=> \(a=\frac{-15\pm\sqrt{65}}{4}\)
Thay vào tính x nhưng bài này tớ ra nghiệm không đẹp
Giải các phương trình sau:
a)\(\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-2}{4-x^2}\)
b)x3- 5x2+ 6x = 0
c)x2+2x-15=0
d)/2x-3/= 4
e)/3x-1/ - x = 2
f)/x-7/ = 2x+3
g)(x+1)4 + (x-3)4 = 82
h)(x+2)(x+3)(x-5)(x-6)= 180
i)\(\frac{x^2-8}{92}+\frac{x^2-7}{93}=\frac{x^2-6}{94}+\frac{x^2-5}{95}\)
1) Giải các pt:
a) 3(x - 1) - 2(x + 3)= -15
b) 3(x - 1) + 2= 3x - 1
c) 7(2 - 5x) - 5= 4(4 -6x)
2) Giải các pt phân thức: ( Tìm mẫu chung )
a) \(\frac{x}{30}+\frac{5x-1}{10}=\frac{x-8}{15}-\frac{2x+3}{6}\)
b) \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
a, Ta có : \(3\left(x-1\right)-2\left(x+3\right)=-15\)
=> \(3x-3-2x-6=-15\)
=> \(3x-3-2x-6+15=0\)
=> \(x=-6\)
Vậy phương trình có nghiệm là x = -6 .
b, Ta có : \(3\left(x-1\right)+2=3x-1\)
=> \(3x-3+2=3x-1\)
=> \(3x-3+2-3x+1=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
c, Ta có : \(7\left(2-5x\right)-5=4\left(4-6x\right)\)
=> \(14-35x-5=16-24x\)
=> \(14-35x-5-16+24x=0\)
=> \(-35x+24x=7\)
=> \(x=\frac{-7}{11}\)
Vậy phương trình có nghiệm là \(x=\frac{-7}{11}\) .
Bài 2 :
a, Ta có : \(\frac{x}{30}+\frac{5x-1}{10}=\frac{x-8}{15}-\frac{2x+3}{6}\)
=> \(\frac{x}{30}+\frac{3\left(5x-1\right)}{30}=\frac{2\left(x-8\right)}{30}-\frac{5\left(2x+3\right)}{30}\)
=> \(x+3\left(5x-1\right)=2\left(x-8\right)-5\left(2x+3\right)\)
=> \(x+15x-3=2x-16-10x-15\)
=> \(x+15x-3-2x+16+10x+15=0\)
=> \(24x+28=0\)
=> \(x=\frac{-28}{24}=\frac{-7}{6}\)
Vậy phương trình có nghiệm là \(x=\frac{-7}{6}\) .
b, Ta có : \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
=> \(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)
=> \(6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)
=> \(6x+24-30x+120=10x-15x+30\)
=> \(6x+24-30x+120-10x+15x-30=0\)
=> \(-19x+114=0\)
=> \(x=\frac{-114}{-19}=6\)
Vậy phương trình có nghiệm là x = 6 .
Câu 3: Giải các phương trình sau bằng cách đưa về dạng ax+b=0
1. a, \(\frac{5x-2}{3}=\frac{5-3x}{2}\); b, \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
c, \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\); d, \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
e, \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\); f, 4 (0,5-1,5x)=\(\frac{5x-6}{3}\)
g, \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\); h, \(\frac{x+4}{5}.x+4=\frac{x}{3}-\frac{x-2}{2}\)
i, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\); k, \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)
m, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\); n, \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right).\frac{1}{3}\left(x+2\right)\)
p, \(\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x\); q, \(\frac{2+x}{5}-0,5x=\frac{1-2x}{4}+0,25\)
r, \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\); s, \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{6}\)
t, \(\frac{2x-8}{6}.\frac{3x+1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\); u, \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{3}+\frac{2x-1}{12}\)
v, \(\frac{5x-1}{10}+\frac{2x+3}{6}=\frac{x-8}{15}-\frac{x}{30}\); w, \(\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x\frac{x-3}{2}}{5}-x+1\)
Đây là những bài cơ bản mà bạn!
\(\frac{5x-2}{3}=\frac{5-3x}{2}\)
\(< =>\frac{\left(5x-2\right).2}{6}=\frac{\left(5-3x\right).3}{6}\)
\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)
\(< =>10x-4=15-9x\)
\(< =>10x+9x=15+4\)
\(< =>19x=19< =>x=1\)
\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)
\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)
\(< =>30x+9=36+24+32x\)
\(< =>32x-30x=9-36-24\)
\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)
Bài 2. Giải các phương trình sau
a, \(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
b, \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
c, \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
d,\(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
e, \(\frac{3\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\left(x-7\right)\)
f, \(\frac{x+5}{2}+\frac{3-2x}{4}=x-\frac{7+x}{6}\)
g, \(\frac{x-3}{11}+\frac{x+1}{3}=\frac{x+7}{9}-1\)
h, \(\frac{3x-0,4}{2}+\frac{1,5-2x}{3}=\frac{x+0,5}{5}\)
a)
\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11-3x+60}{12}=0\)
\(\Leftrightarrow\frac{49-13x}{12}=0\)
\(\Rightarrow49-13x=0\)
\(\Rightarrow x=\frac{-49}{13}\)
b)
\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)
\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)
\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)
\(\Leftrightarrow\frac{-3x}{4}=0\)
\(\Rightarrow-3x=0\)
\(\Rightarrow x=0\)
e)
\(\frac{3\cdot\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\cdot\left(x-7\right)\)
\(\Leftrightarrow\frac{45x-18-24-28x+60x-420}{12}=0\)
\(\Leftrightarrow\frac{77x-462}{12}=0\)
\(\Rightarrow77x-462=0\)
\(\Rightarrow x=\frac{462}{77}=6\)
tìm x biết
a) 5x(2x-7)+2x(8-5x)=5
b)(3x-5)(7-5x)-(5x+2)(2-3x)=4
c)5(2x-3)^2-5(x+1)^2-15(x+4)(x-4)=-10
d) 5x(x-3)(x+3)-(2x-3)^2 -5(x+2)^3+34x(x+2)=1
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
giải các pt sau
a)5X(X-2020)+X=2020
b)4(X-5)2-(2X+1)2=0
c)\(\frac{3X}{5}-\frac{2X+1}{3}=2-\frac{X-3}{15}\)
d)5X3+10X2+5X=0
e)2X3-8X=0
f)\(\frac{X^2+5}{25-X^2}=\frac{3}{X+5}+\frac{X}{X-5}\)
g)\(\frac{4}{2X-3}-\frac{4X}{9-4X^2}=\frac{1}{2X+3}\)
h)|2X-4|-15=1
i)20-3|2X+1|=17
k)|4X+2|-1,5=1
GIẢI GIÚP MÌNH NHANH VỚI NHA
\(5X\left(X-2020\right)+X=2020\)
\(\Leftrightarrow5X^2-10100X+X=2020\)
\(\Leftrightarrow5X^2-10099X=2020\)
\(\Leftrightarrow5X^2-10099X-2020=0\)
\(\Leftrightarrow5X^2-10100X+x-2020=0\)
\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)
\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)
\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)
\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)
\(\Leftrightarrow-11\left(4x-9\right)=0\)
\(\Leftrightarrow x=\frac{9}{4}\)
\(a,5x\left(x-2020\right)+x=2020\)
\(< =>5x\left(x-2020\right)+x-2020=0\)
\(< =>\left(5x+1\right)\left(x-2020\right)=0\)
\(< =>\orbr{\begin{cases}5x+1=0\\x-2020=0\end{cases}}\)
\(< =>\orbr{\begin{cases}5x=-1\\x=2020\end{cases}< =>\orbr{\begin{cases}x=-\frac{1}{5}\\x=2020\end{cases}}}\)
\(b,4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(< =>4\left(x^2-20x+25\right)-\left(4x^2+4x+1\right)=0\)
\(< =>4x^2-80x+100-4x^2-4x-1=0\)
\(< =>-84x+99=0< =>84x=99< =>x=\frac{99}{84}\)
đọc cách lm trrong sbt nha bạn lớp 8
mk học lớp 6 lên 7
Tìm x
a, \(-\frac{22}{15}x+\frac{1}{3}=\text{|}-\frac{2}{3}+\frac{1}{5}\text{|}\)
b,\(x:15=8:24\)
c,\(\frac{-3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
d, \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)
nói cách làm lun nha
a, \(-\frac{22}{15}x+\frac{1}{3}=\left|-\frac{2}{3}+\frac{1}{5}\right|=\left|-\frac{7}{15}\right|=\frac{7}{15}\)
\(\Rightarrow\frac{-22}{15}x=\frac{7}{15}-\frac{1}{3}=\frac{2}{15}\)
\(\Rightarrow x=\frac{2}{15}:\frac{-22}{15}=\frac{2}{15}.\frac{15}{-22}=-\frac{1}{11}\)
b,\(x:15=8:24\)
Vậy x=5
\(\Rightarrow x:15=\frac{8}{24}\Rightarrow x=\frac{8}{24}.15=\frac{120}{24}=5\)