a. 7x – 8 = 4 x + 7
b. \(\dfrac{2x-1}{3}\)- x+2=\(\dfrac{3x+5}{4}\)
Thu gọn các đơn thức đồng dạng trong đa thức sau:
a) \(A=x^5-\dfrac{1}{2}x+7x^3-2x+\dfrac{1}{5}x^3+3x^4-x^5+\dfrac{2}{5}x^4+15\)
b) \(B=3x^2-10+\dfrac{2}{5}x^3+7x-x^2+8+7x^2\)
c) \(C=\dfrac{1}{7}x-2x^4+5x+6\)
Thu gọn các đơn thức đồng dạng trong đa thức sau:
a) \(A=x^5-\dfrac{1}{2}x+7x^3-2x+\dfrac{1}{5}x^3+3x^4-x^5+\dfrac{2}{5}x^4+15\)
b) \(B=3x^2-10+\dfrac{2}{5}x^3+7x-x^2+8+7x^2\)
c) \(C=\dfrac{1}{7}x-2x^4+5x+6\)
a)A=\(x^5-\dfrac{1}{2}x+7x^3-2x+\dfrac{1}{5}x^3+3x^4-x^5+\dfrac{2}{5}x^4+15\)
=\(=\dfrac{-5}{2}x+\dfrac{36}{5}x^3+\dfrac{17}{5}x^4+15\)
b)B=\(3x^2-10+\dfrac{2}{5}x^3+7x-x^2+8+7x^2\)
\(=9x^2+\dfrac{2}{5}x^3+7x+2\)
c)C=\(\dfrac{1}{7}x-2x^4+5x+6\)
1. \(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)
2. \(x-\dfrac{3\left(x+30\right)}{15}-24\dfrac{1}{2}=\dfrac{7x}{10}-\dfrac{2\left(10x+2\right)}{5}\)
3. \(14\dfrac{1}{2}-\dfrac{2\left(x+3\right)}{5}=\dfrac{3x}{2}-\dfrac{2\left(x-7\right)}{3}\)
4. \(\dfrac{x+1}{3}+\dfrac{3\left(2x+1\right)}{4}=\dfrac{2x+3\left(x+1\right)}{6}+\dfrac{7+12x}{12}\)
5. \(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
6. \(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)
7. \(\dfrac{3\left(x-3\right)}{4}+\dfrac{4x-10,5}{10}=\dfrac{3\left(x+1\right)}{5}+6\)
8. \(\dfrac{2\left(3x+1\right)+1}{4}-5=\dfrac{2\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\)
1) giải phương trình :
a) \(\left(2+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
b) \(\dfrac{7x+10}{x+1}\left(x^2-x-2\right)-\dfrac{7x+10}{x+1}\left(2x^2-3x-5\right)=0\)
c) \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
d) \(\dfrac{13}{2x^2+x-21}+\dfrac{1}{2x+7}+\dfrac{6}{9-x^2}=0\)
i) \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
k) \(\dfrac{1+\dfrac{x}{x+3}}{1-\dfrac{x}{x+3}}=3\)
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
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, \(\Rightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-105.17\)
\(\Leftrightarrow70x-105-30x-45=84x+63-1785\)
\(\Leftrightarrow-44x=-1572\Leftrightarrow x=\dfrac{393}{11}\)
1/ \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
2/ \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
3/ \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
4/ \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
5/ \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
1).(4-3x)(10-5x)=0 2).(7-2x)(4+8x)=0 3).(9-7x)(11-3x)=0
4).(7-14x)(x-2)=0 5).(\(\dfrac{7}{8}\)-2x)(3x+\(\dfrac{1}{3}\))=0 6).3x-2x\(^2\)
7).5x+10x\(^2\)
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
a) (X-2)(x+3)-3(4x-2)=(x-4)\(^{^{ }2}\)
b) \(\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\)
c) \(x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
d) \(\left(2x+5\right)^2=\left(x+2\right)^2\)
e) \(x^2-5+6=0\)
g) \(2x^3+6x^2=x^2+3x\)
h) \(\left(x+\dfrac{1}{2}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\)
mọi người giúp e với ạ
\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)
\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)
\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)
\(d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)
Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:
\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)
Tick plzz
a: Ta có: \(\left(x-2\right)\left(x+3\right)-3\left(4x-2\right)=\left(x-4\right)^2\)
\(\Leftrightarrow x^2+x-6-12x+6-x^2+8x-16=0\)
\(\Leftrightarrow-3x=16\)
hay \(x=-\dfrac{16}{3}\)
b: Ta có: \(\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\)
\(\Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\)
\(\Leftrightarrow-14x+7+4x-6=0\)
\(\Leftrightarrow10x=1\)
hay \(x=\dfrac{1}{10}\)
c: Ta có: \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)
\(\Leftrightarrow23x+70=10x+200\)
\(\Leftrightarrow x=10\)
Tìm x
\(a,3-x=x+1,8\)
\(b,2x-5=7x+35\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(k,7x^2-11=6x^2-2\)
\(m,5\left(x+3.2^3\right)=10^2\)
\(n,\dfrac{4}{9}-(\dfrac{1}{6^2})=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
a: 3-x=x+1,8
=>-2x=-1,2
=>x=0,6
b: 2x-5=7x+35
=>-5x=40
=>x=-8
c: 2(x+10)=3(x-6)
=>3x-18=2x+20
=>x=38
d; 8(x-3/8)+1=6(1/6+x)+x
=>8x-3+1=1+6x+x
=>8x-2=7x+1
=>x=3
e: =>-3x+x=4/3-2/9
=>-2x=12/9-2/9=10/9
=>x=-5/9
g: =>3/4x-1/2x=5/6+1/2
=>1/4x=5/6+3/6=8/6=4/3
=>x=4/3*4=16/3
h: =>x-4=-x+5
=>2x=9
=>x=9/2