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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
\(d,=24x^2-38x+3\\ e,=x^2-12x+35\\ f,=\left(x^2-144\right)\left(4x-1\right)=4x^3-x^2-576x+144\)
a/ \(\dfrac{3-x}{12}=\dfrac{2x+2}{8}\)
\(< =>\dfrac{2\left(3-x\right)}{24}=\dfrac{3\left(2x+2\right)}{24}\)
\(< =>6-2x-6x-6=0\)
\(< =>-8x=0\)
\(< =>x=0\)
Vậy tập nghiệm.....
b/ \(\dfrac{x+3}{x-4}+\dfrac{x-3}{x+4}=\dfrac{2\left(x^2+12\right)}{x^2-16}\)
Tìm ĐKXĐ của pt là: \(x\ne\pm4\) (làm tắt, bạn làm rõ ra nhé)
\(\dfrac{x+3}{x-4}+\dfrac{x-3}{x+4}=\dfrac{2\left(x^2+12\right)}{x^2-16}\)
\(< =>\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{2\left(x^2+12\right)}{\left(x+4\right)\left(x-4\right)}\)
\(< =>x^2+3x+4x+12+x^2-3x-4x+12-2x^2-24=0\)
\(< =>0x=0\)
=> x có vô số nghiệm
Vậy ....
a) `(3-x)/12=(2x+2)/8`
`<=> (3-x)/12 =(x+1)/4`
`<=> 3-x=3(x+1)`
`<=>3-x=3x+3`
`<=> x=0`
Vậy `S={0}`.
b) ĐK: `x \ne \pm 4`
`(x+3)/(x-4)+(x-3)/(x+4)=(2(x^2+12))/(x^2-16)`
`<=> (x+3)(x+4)+(x-3)(x-4)=2(x^2+12)`
`<=> x^2+7x+12+x^2-7x+12=2x^2+24`
`<=> 0x=0`
Vậy PT có nghiệm với mọi x thỏa mãn điều kiện.
\(\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}=\dfrac{x\left(x+12\right)}{x^2-9}\) \(\left(x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{\left(x-3\right)^2-\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x\left(x+12\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x+3\right)^2=x\left(x+12\right)\)
\(\Leftrightarrow x^2-6x+9-x^2-6x-9=x^2+12x\)
\(\Leftrightarrow-12x=x^2+12x\)
\(\Leftrightarrow x^2+24x=0\)
\(\Leftrightarrow x\left(x+24\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-24\end{matrix}\right.\) (n)
Vậy \(S=\left\{-24;0\right\}\)
=>(x-3)(x-3)-(x+3)(x+3)=x^2+12x
=>x^2-6x+9-x^2-6x-9=x^2+12x
=>x^2+12x=-12x
=>x^2+24x=0
=>x=0 hoặc x=-24
a) \(10x-\frac{5}{18}+x+\frac{3}{12}=7x+\frac{3}{6}-12-\frac{x}{9}\)
\(10x+x-7x+\frac{x}{9}=\frac{3}{6}+\frac{5}{18}-\frac{3}{12}-12\)
\(\frac{37}{9}x=-\frac{413}{36}\)
\(x=-\frac{413}{148}\)
=) vào ngay quả bảng phá dấu GTTĐ, cay thế :<
a, \(3x+\frac{2x}{3}-3=\frac{5}{2}x-2\Leftrightarrow\frac{18x+4x-18}{6}=\frac{15x-12}{6}\)
\(\Rightarrow22x-18=15x-12\Leftrightarrow7x=6\Leftrightarrow x=\frac{6}{7}\)
Vậy pt có nghiệm x = 6/7
b, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)
\(\Leftrightarrow\frac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\frac{x+7}{12}\)
\(\Rightarrow18x+9-10x-6+4x+4=x+7\)
\(\Leftrightarrow12x+7=x+7\Leftrightarrow11x=0\Leftrightarrow x=0\)
Vậy pt có nghiệm là x = 0
c, \(\frac{3x}{x-3}-\frac{x-3}{x+3}=2\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow\frac{3x\left(x+3\right)-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow3x^2+9x-x^2+6x-9=2\left(x^2-9\right)\)
\(\Leftrightarrow2x^2+15x-9=2x^2-18\Leftrightarrow15x+9=0\Leftrightarrow x=-\frac{9}{15}=-\frac{3}{5}\)
Vậy pt có nghiệm là x = -3/5
d, Sửa đề : \(\frac{x+10}{2003}+\frac{x+6}{2007}+\frac{x+2}{2011}+3=0\)
\(\Leftrightarrow\frac{x+10}{2003}+1+\frac{x+6}{2007}+1+\frac{x+2}{2011}+1=0\)
\(\Leftrightarrow\frac{x+2013}{2003}+\frac{x+2013}{2007}+\frac{x+2013}{2011}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2003}+\frac{1}{2007}+\frac{1}{2011}\ne0\right)=0\Leftrightarrow x=-2013\)
Vậy pt có nghiệm là x = -2013
e, \(4\left(x+5\right)-3\left|2x-1\right|=10\)
\(\Leftrightarrow4x+20-3\left|2x-1\right|=10\Leftrightarrow-3\left|2x-1\right|=-10-4x\)
\(\Leftrightarrow\left|2x-1\right|=\frac{10+4x}{3}\)
ĐK : \(\frac{10+4x}{3}\ge0\Leftrightarrow10+4x\ge0\Leftrightarrow x\ge-\frac{10}{4}=-\frac{5}{2}\)
TH1 : \(2x-1=\frac{10+4x}{3}\Rightarrow6x-3=10+4x\Leftrightarrow2x=13\Leftrightarrow x=\frac{13}{2}\)( tm )
TH2 : \(2x-1=\frac{-10-4x}{3}\Rightarrow6x-3=-10-4x\Leftrightarrow10x=-7\Leftrightarrow x=-\frac{7}{10}\)( tm )
f, để mình xem lại đã, quên cách phá GTTĐ rồi :v :>
k,\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\)
giúp mk câu k nhé đề bài như trên
b: \(\Leftrightarrow4x+8-9=4x-4\)
=>-1=-4(loại)
d: \(\Leftrightarrow3\left(x-2\right)+2\left(x+1\right)=8x\)
=>8x=3x-6+2x+2=5x-4
=>3x=-4
=>x=-4/3
f: \(\Leftrightarrow3\left(x+2\right)+4\left(2x-3\right)=2\left(x-12\right)\)
=>3x+6+8x-12=2x-24
=>11x-6=2x-24
=>9x=-18
=>x=-2
Ta có : x4 + x3 + 6x2 + 5x + 5
= (x4 + 5x2) + (x3 + 5x) + (x2 + 5)
= x2(x2 + 5) + x(x2 + 5) + (x2 + 5)
= (x2 + 5)(x2 + x + 1)
Có 2 trường hợp xảy ra
TH1: x=3
TH2: x=-3
#Hk_tốt
#Ngọc's_Ken'z
\(\left|x-3\right|+\left|3-x\right|=12\)
\(\Leftrightarrow\left|x-3\right|+\left|x-3\right|=12\)
\(\Leftrightarrow2\left|x-3\right|=12\)
\(\Leftrightarrow\left|x-3\right|=6\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=6\\x-3=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-3\end{cases}}\)
Vậy....