1 Tìm x
a.11/12 - ( 2/5 +x ) = 2/3
b. 3/4 + 1/4 : x = 2/5
c. ( 3/2 - x) - 1/ -3 = 3/4
d. 3/4x - 1/2 = 5/3 + 1/2 x
Chú ý : Đây là phân số
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)
\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)
Suy ra: \(-12x-3=8x-2-6x-8\)
\(\Leftrightarrow-12x-3-2x+10=0\)
\(\Leftrightarrow-14x+7=0\)
\(\Leftrightarrow-14x=-7\)
\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
a)\(\dfrac{2}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{4}{5}=\left(\dfrac{2}{3}+\dfrac{1}{3}\right).\dfrac{4}{5}=1.\dfrac{4}{5}=\dfrac{4}{5}\)
b)\(\dfrac{2}{3}.\dfrac{4}{5}-\dfrac{1}{3}.\dfrac{4}{5}=\left(\dfrac{2}{3}-\dfrac{1}{3}\right).\dfrac{4}{5}=\dfrac{1}{3}.\dfrac{4}{5}=\dfrac{4}{15}\)
a) \(\dfrac{2}{3}\times\dfrac{4}{5}+\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=\dfrac{4}{5}\times1=\dfrac{4}{5}\)
b) \(\dfrac{2}{3}\times\dfrac{4}{5}-\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\left(\dfrac{2}{3}-\dfrac{1}{3}\right)=\dfrac{4}{5}\times\dfrac{1}{3}=\dfrac{4}{15}\)
c) \(\dfrac{1}{2}:\dfrac{3}{4}+\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}+\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\times\left(\dfrac{1}{2}+\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{2}{3}=\dfrac{8}{9}\)
d) \(\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}-\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{1}{3}=\dfrac{4}{9}\)
a: =>x/27+1=-2/3
=>x/27=-5/3
=>x=-45
b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)
=>x=221/50
c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)
=>x=1/15-2/3=1/15-10/15=-9/15=-3/5
d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)
=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)
=>x=-37/24
e: =>-3/7x=84/45
=>x=-196/45
f: =>11/10x=-2/3
=>x=-20/33
a: \(\Leftrightarrow\left|x\cdot\dfrac{7}{3}-\dfrac{3}{4}\right|=1+\dfrac{1}{3}+\dfrac{2}{3}=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{7}{3}-\dfrac{3}{4}=2\\x\cdot\dfrac{7}{3}-\dfrac{3}{4}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{33}{28}\\x=-\dfrac{15}{28}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|x\cdot\dfrac{2}{3}-\dfrac{1}{3}\right|=\dfrac{6}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{6}{5}\\x\cdot\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-13}{10}\\x=\dfrac{23}{10}\end{matrix}\right.\)
2: \(=\dfrac{-2}{75}+\dfrac{5}{39}=\dfrac{33}{325}\)
3: \(=\dfrac{6}{11}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)=\dfrac{6}{11}\)
4: \(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-1\right)=-2\cdot\dfrac{7}{19}=-\dfrac{14}{19}\)
5: \(=\dfrac{2}{7}\left(\dfrac{4}{23}-\dfrac{27}{23}+1\right)=0\)
6: \(=\dfrac{3}{8}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\dfrac{11}{8}=\dfrac{3}{8}+\dfrac{11}{8}=\dfrac{14}{8}=\dfrac{7}{4}\)
1,
Từ đề bài => a/c * c/b = (a/c)^2=(c/b)^2
=> a/b=a^2/c^2=c^2/b^2=a^2+c^2/c^2+b^2=> a/b=a^2+c^2/c^2+b^2
=> DPCM
(từ mình làm tiếp)
Bài này mink làm trc
2,
Đặt a/b=c/d=k
=> a=kb, c=kd
Ta có:
5a+3b/5a-3b=5kb+3b/5kb-3b
=3b*(2k+1)/3b*(2k-2)=2k+1/2k-1
Chứng minh tương tự với biểu thức 5c+3d/5c-3d
Ta cũng đc 2k+1/2k-1
=> Nếu a/b=c/d thì 5a+3b/5a-3b=5c+3d/5a-3d
=> dpcm
\(a,2\left(x^3-1\right)-2x^2\left(x+2x^4\right)+x\left(4x^5+4\right)=6\\ \Leftrightarrow2x^3-2-2x^3-4x^6+4x^6+4x-6=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow x=2\\ b,\left(2x\right)^2\left(4x-2\right)-\left(x^3-8x^3\right)=15\\ \Leftrightarrow4x^2\left(4x-2\right)+7x^3-15=0\\ \Leftrightarrow16x^3-8x^2+7x^3-15=0\\ \Leftrightarrow23x^3-8x^2-15=0\\ \Leftrightarrow23x^3-23x^2+15x^2-15x+15x-15=0\\ \Leftrightarrow\left(x-1\right)\left(23x^2+15x-15\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x\in\varnothing\left(23x^2+15x-15>0\right)\end{matrix}\right.\)
Bài 1:
a: Ta có: \(2\left(x^3-1\right)-2x^2\left(2x^4+x\right)+x\left(4x^5+4\right)=6\)
\(\Leftrightarrow2x^3-2-4x^6-2x^3+4x^6+4x=6\)
\(\Leftrightarrow4x=8\)
hay x=2
b: Ta có: \(\left(2x\right)^2\cdot\left(4x-2\right)-\left(x^3-8x^3\right)=15\)
\(\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^3=15\)
\(\Leftrightarrow16x^3-8x^2+7x^3=15\)
\(\Leftrightarrow23x^3-8x^2-15=0\)
\(\Leftrightarrow23x^3-23x^2+15x^2-15=0\)
\(\Leftrightarrow23x^2\left(x-1\right)+15\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(23X^2+15x+15\right)=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
a,
Ta có:
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{55}{60}-\frac{60\left(\frac{2}{5}+x\right)}{60}=\frac{40}{60}\)
\(\Leftrightarrow\frac{55}{60}-\frac{24+60x}{60}=\frac{40}{60}\)
=> 55-24-60x=40
<=> 31-60x=40
<=> x=-3/20
Mấy câu còn lại cũng tương tự thế đó