Chỉ câu này giúp mình với:
1 tìm x:
9x56xxx2486146-9594599565=669xx925
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Làm thế này nhé
(2x-6)y+x-3=18
(2x-6)y+x-18-3=0
(2x-6)y+x-21=0
sau đó bạn rút gọn nhé, ta làm như sau
2(x-3)=0
=> 2x= 2 x 3
=> x=3 nhé
giúp mình llàm câu b với nhé đã k cho bạn rồi nha
b, y = \(\frac{23}{x+7}\)
ĐỀ:
\(\dfrac{2}{x}-2+41=\dfrac{7}{2}\)
ĐÁP ÁN:
\(x=\dfrac{-4}{71}\)
\(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-2=0\)
\(\Rightarrow x^3+3x^2+3x+1-x^3+1-2=0\)
\(\Rightarrow3x^2+3x=0\Rightarrow3x\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Ta có :
\(\frac{x-3}{97}+\frac{x-27}{73}+\frac{x-67}{33}+\frac{x-73}{27}=4\)
\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)
\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)
Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}>0\) Nên \(x-100=0\)
\(\Leftrightarrow x=100\)
Vậy \(x=100\)
\(\Leftrightarrow\frac{x-3}{87}+\frac{x-27}{79}+\frac{x-67}{33}+\frac{x-73}{27}-4=0\)
\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)
\(\Leftrightarrow\left(\frac{x-3-97}{97}\right)+\left(\frac{x-27-73}{73}\right)+\left(\frac{x-67-33}{33}\right)+\left(\frac{x-73-27}{27}\right)=0\)
\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)
Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\ne0\)
\(\Rightarrow x-100=0\Leftrightarrow x=100\)
\(2x-49=5.32\\ \Leftrightarrow2x-49=160\\ \Leftrightarrow2x=209\\ \Leftrightarrow x=\dfrac{209}{2}\)
\(200-\left(2x+6\right)=43\\ \Leftrightarrow2x+6=157\\ \Leftrightarrow2x=151\\ \Leftrightarrow x=\dfrac{151}{2}\)
\(135-5\left(x+4\right)=35\\ \Leftrightarrow5\left(x+4\right)=100\\ \Leftrightarrow x+4=20\\ \Leftrightarrow x=16\)
Ta có: lx-1l + l4-xl = 3 <=> lx-1l + lx-4l = 3
TH1: Nếu x < 1, ta có: TH2: Nếu 1 < x < 4, ta có: TH3: Nếu x > 4, ta có: 1 - x + 4 - x = 3 x - 1 + 4 - x = 3 x - 1 + x - 4 = 3 <=>5 - 2x = 3 <=> 3 =3 (TM) <=> 2x - 5 = 3
<=> 2x = 5 - 3 = 2 <=> x = 1;2;3;4 <=> 2x = 3 + 5 = 8 <=> x = 1 (TM) < => x = 4(TM) Vậy x = 1;2;3;4.
chỉ giúp mình