\(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\)
3x2-7x+1=0
2x2-6x+1=0
mọi người giúp em với
Giải phương trình:
1) \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
2) \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
3)\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
4)\(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
5)\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
6)\(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)
7)\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
8)\(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
giải giúp mik với ạ
1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow95x-5=96-6x\)
\(\Leftrightarrow95x+6x=96+5\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9=32x+60\)
\(\Leftrightarrow30x-32x=60-9\)
\(\Leftrightarrow-2x=51\)
\(\Leftrightarrow x=-\frac{51}{2}\)
3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)
=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)
=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)
=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)
=> 27 - 9x + 80 - 16x = 12 - 12x - 48
=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0
=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0
=> 143 - 13x = 0
=> 13x = 143
=> x = 11
5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)
=> 6x - 18 + 7x - 35 - 13x - 4 = 0
=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0
=> -57 = 0(vô nghiệm)
6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)
=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)
=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)
=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)
=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)
=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)
=> \(12x+10-10x-3=12x+2\)
=> \(2x+10-3=12x+2\)
=> 2x + 10 - 3 - 12x - 2 = 0
=> (2x - 12x) + (10 - 3 - 2) = 0
=> -10x + 5 = 0
=> -10x = -5
=> x = 1/2
7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)
=> 6x - 3 - 5x + 10 - x - 7 = 0
=> (6x - 5x - x) + (-3 + 10 - 7) = 0
=> 0x + 0 = 0
=> 0x = 0
=> x tùy ý
Bài 8 tự làm nhé
Xin mọi người hãy giúp em giải bài toán này! Em còn rất ít thời gian nữa . Em xin chân thành cảm ơn!
a, \(7^{2x}+7^{2x+2}=2450\)
b, \(33^x:11^x=81\)
c, \(\frac{5}{3}:10=\frac{2}{3}:\left(8x\right)\)
d, \(\frac{\left(x+1\right)^2}{8}=\frac{8}{x+1}\)với x khác 1
e, \(x^2-7x+12< 0\)
g, \(\left(x-\frac{1}{2}\right)\times\left(x+5\right)>0\)
Xin mọi người hãy giúp em ! Em xin chân thành cảm ơn !
a)72x+72x.49=2450
72x.50=2450
72x=2450:50=49
72x=72
2x=2
x=1
b)(33:11)x=81
3x=81
3x=34
x=4
c)1/6=2/3:8x
8x=2/3:1/6
8x=4
x=1/2
d)(x+1)3=64
(x+1)3=43
x+1=4
x=3
minh chỉ lam đc vậy thôi nha !hi hi
Giái phương trình :
a,\(\frac{6x+1}{x^2-7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)
b,\(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)
c,\(\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{x^2-2x-3}\)
d,\(\frac{2}{x+2}-\frac{2x^2+16}{x^3+8}=\frac{5}{x^2-2x+4}\)
a) ĐKXĐ: x∉{2;5}
Ta có: \(\frac{6x+1}{x^2-7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)
\(\Leftrightarrow\frac{6x+1}{\left(x-2\right)\left(x-5\right)}+\frac{5\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{3\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}=0\)
\(\Leftrightarrow6x+1+5x-25-3\left(x-2\right)=0\)
\(\Leftrightarrow11x-24-3x+6=0\)
\(\Leftrightarrow8x-18=0\)
\(\Leftrightarrow8x=18\)
hay \(x=\frac{9}{4}\)(tm)
Vậy: \(x=\frac{9}{4}\)
b) ĐKXĐ: x∉{0;2;-2}
Ta có: \(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x}{x\left(x-2\right)\left(x+2\right)}-\frac{\left(x-1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x-4\right)\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}=0\)
\(\Leftrightarrow2x-\left(x^2+x-2\right)+x^2-6x+8=0\)
\(\Leftrightarrow2x-x^2-x+2+x^2-6x+8=0\)
\(\Leftrightarrow-5x+10=0\)
\(\Leftrightarrow-5x=-10\)
hay x=2(ktm)
Vậy: x∈∅
Giải các phương trình sau:
\(\frac{3}{4x-20}-\frac{15}{2x^2-50}+\frac{7}{6x+30}=0\)
\(\frac{8x^2}{3-12x^2}+\frac{1+8x}{4+8x}=\frac{-2x}{3-6x}\)
\(\frac{1}{x^2-2x+1}+\frac{1}{x^2+2x=1}=\frac{2}{x^2-1}\)
\(\frac{1}{x^2+1}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}=\frac{4}{5}\)
Tìm x
1) \(\frac{-1}{3}\)x.(1+x)+x.(\(\frac{1}{3}\)x-2)=5
2)\(^{2x2}\)-7x+5=0
1) Ta có: \(\frac{-1}{3}x\left(1+x\right)+x\left(\frac{1}{3}x-2\right)=5\)
\(\Leftrightarrow\frac{-1}{3}x+\frac{-1}{3}x^2+\frac{1}{3}x^2-2x=5\)
\(\Leftrightarrow\frac{-7}{3}x=5\)
\(\Leftrightarrow x=5:\frac{-7}{3}=\frac{15}{-7}\)
Vậy: \(x=-\frac{15}{7}\)
2) Ta có: \(2x^2-7x+5=0\)
\(\Leftrightarrow2x^2-2x-5x+5=0\)
\(\Leftrightarrow2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{2;\frac{5}{2}\right\}\)
giải phương trình
a/ x^4-3x^3+6x^2-3x+1=0
b/ (4x+3)^3+(5-7x)^3+(3x-8)^3=0
c/ (x-2)^4 +(x-3)^4=1
d/ x^2 + \(\frac{1}{x^2}\)+ y^2 + \(\frac{1}{y^2}\)=4
a)
voi x=0 ta thay 0 o phai la no pt
voi x<>0 chia ca 2 ve cho x^2 ta dc
x^2-3x+6-3/x+1/x^2=0
(x^2+1/x^2)-3(x+1/x)+6=0 dat a=x+1/x ta co (x+1/x)^2=a^2=>x^2+1/x^2=a^2-2
=>a^2-3a+4=0=>pt vo no :(
giải phương trình
a/ x^4-3x^3+6x^2-3x+1=0
b/ (4x+3)^3+(5-7x)^3+(3x-8)^3=0
c/ (x-2)^4 +(x-3)^4=1
d/ x^2 + \(\frac{1}{x^2}\)+ y^2 + \(\frac{1}{y^2}\)=4
Giải phương trình:
1.\(\frac{x-5}{x-5}+\frac{x-6}{x-5}+\frac{x-7}{x-5}+...+\frac{1}{x-5}=4\left(x\in N\right)\)
2.\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)
3.\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{x\left(x+2\right)}\right)=\frac{31}{16}\left(x\in N\right)\)
4.\(8\left(x^2+\frac{1}{x^2}\right)-34\left(x+\frac{1}{x}\right)+51=0\)
5.\(6x^4-5x^3-38x^2-5x+6=0\)
Cho E = \(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)\) : \(\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-2}\right)\)
a. Rút gọn E
b. Tính E khi x² - 9 = 0
c. Tìm giá trị của x để E = 3
d. Tìm x để E<0
e. Tính x khi E - x - 3 = 0
Mọi người giúp em với ạ. Xin cảm ơn.