Giup minh voi 1/3x+2/5(x+1)=0
LH
Những câu hỏi liên quan
a,|x-3| - | 2x-4| = 0
b, |3x-2| = x-1
c, |4-3x| = 2x+1
d, |4-2x| +|x+1| =6
giup minh voi
Minh can gap giup minh voi:
a) \3x=9\=\4-x\+0
b)(x-5)16 =(x-5)12
c)(2x-5)(5-x mu 2)(x mu 2 +1=0
d)5 mu x + 5 mu x+2 =650
Câu này khó quá nên mình không làm được
(3x+1\5).(x-1\2)=0
(X-3\2).(2X+1)>0
(2-x)-(4\5-x)<0
giup mk voi nhe
(x^2+x+1).3x+1/x+2=(x^2+x+1).x/2(x+2) Giup minh voi
\(\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}=\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}\) \(\left(dkxd:x\ne-2\right)\)
\(\Leftrightarrow\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}-\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left[2\left(3x+1\right)-x\right]=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(6x+2-x\right)=0\)
Bỏ vế đằng trước \(x^2+x+1=0\) do vô nghiệm
\(\Leftrightarrow6x+2-x=0\)
\(\Leftrightarrow5x=-2\)
\(\Leftrightarrow x=-\dfrac{2}{5}\left(tmdk\right)\)
Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
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\(\dfrac{\left(x^2+x+1\right).\left(3x+1\right)}{x+2}=\dfrac{\left(x^2+x+1\right).x}{2\left(x+2\right)}\)
hay \(\left(x^2+x+1\right).\dfrac{3x+1}{x+2}=\left(x^2+x+1\right).\dfrac{x}{2\left(x+2\right)}\)
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A)4X(X-5)+X(X-1)
B)3X(6X-1)-X(X-6)
C)X(X-5)-X(X+3)-6(X+8)
D)Xmu2(X-2)-X(Xmu2-6)
giup minh voi=(
Tim x, biet:
a)(x-3)^3-(x-3)(x^2+3x+9)+9(x-1)^2=15
b)(x^2-2)^2+4(x-1)^2-4(x^2-2)(x-1)=0
Giup minh voi!
1, Tìm X,..
a,4/5.x/-1=5/4
5/9-1/3x=1/2
3/4x-1/4x+7/2=-5/4
5/8+1/8x=-1/2
giup minh voi a
|3x-2018| + |x-2017| = |2x -1|
|x-1| + |x-3| +|x-5| +|x-7| = 8
|x-2018| + |x-2017| + |x-2018| = 2
MOi nguoi giup minh voi 4gio minh can roi
\(\left|3x-2018\right|+\left|x-2017\right|=\left|2x-1\right|\)
\(\Rightarrow\orbr{\begin{cases}3x-2018+x-2017=2x-1\\-\left(3x-2018\right)+\left[-\left(x-2017\right)\right]=2x-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x-4035=2x-1\\\left(-3x-x\right)+\left(2018+2017\right)=2x-1\end{cases}}\)
Làm tiếp
TH2:
\(\left|3x-2018\right|+\left|x-2017\right|=\left|2x-1\right|\)
\(\Rightarrow\orbr{\begin{cases}3x-2018+x-2017=-2x+1\\-\left(3x-2018\right)+\left[-\left(x-2017\right)\right]=-2x+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x-4035=-2x+1\\\left(-3x-x\right)+\left(2018+2017\right)=-2x+1\end{cases}}\)
Tự tiếp tiếp nha bạn
Bài sau cũng tg tự vậy mà làm
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tim x biet: (x-3)*(x-5)+1=0 giup minh voi. Arigato gozaimasu!
\(\left(x-3\right)\left(x-5\right)+1=0\)
\(\Leftrightarrow x^2-5x-3x+15+1=0\)
\(\Leftrightarrow x^2-8x+16=0\)
\(\Leftrightarrow\left(x-4\right)^2=0\)
\(\Leftrightarrow x-4=0\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\)
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\(\left(x-3\right)\left(x-5\right)+1=0\)
\(\Rightarrow x^2-3x-5x+15+1=0\)
\(\Rightarrow x^2-8x+16=0\)
\(\Rightarrow x^2-2x.4+4^2=0\)
\(\Rightarrow\left(x-4\right)^2=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
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\(\left(x-3\right)\left(x-5\right)+1=0\)
\(\Leftrightarrow x^2-5x-3x+15+1=0\)
\(\Leftrightarrow x^2-8x+16=0\)
\(\Leftrightarrow\left(x-4\right)^2=0\)
\(\Leftrightarrow x=4\)
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