Tìm x:
a, 2/7x-1/3=3/5x-1
b, |3x+7|-3x=7
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a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)
\(\Leftrightarrow2x-1=0\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)
\(\Leftrightarrow x^3-x^3-1=x\)
hay x=-1
c: Ta có: \(56x^4+7x=0\)
\(\Leftrightarrow7x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d: Ta có: \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
a. \(x^2-2x+2\left|x-1\right|-7=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-2x+2\left(x-1\right)-7=0\\x^2-2x-2\left(x-1\right)-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-9=0\\x^2-4x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=9\\\left(x-5\right)\left(x+1\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm3\\x=5\\x=-1\end{matrix}\right.\)
b: Ta có: \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)
\(\Leftrightarrow\left(x^2+5x\right)^2+10\cdot\left(x^2+5x\right)=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
\(B=3x^2+3x-1\)
\(=3\left(x^2+x-\dfrac{1}{3}\right)\)
\(=3\left(x^2+x+\dfrac{1}{4}-\dfrac{7}{12}\right)\)
\(=3\left(x+\dfrac{1}{2}\right)^2-\dfrac{7}{4}>=-\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi x+1/2=0
=>\(x=-\dfrac{1}{2}\)
\(C=-2x^2+7x+3\)
\(=-2\left(x^2-\dfrac{7}{2}x-\dfrac{3}{2}\right)\)
\(=-2\left(x^2-2\cdot x\cdot\dfrac{7}{4}+\dfrac{49}{16}-\dfrac{73}{16}\right)\)
\(=-2\left(x-\dfrac{7}{4}\right)^2+\dfrac{73}{8}< =\dfrac{73}{8}\forall x\)
Dấu '=' xảy ra khi x-7/4=0
=>x=7/4
a) 6x(5x + 3) + 3x(1 – 10x) = 7
⇒ 30x2+18x+3x-30x2=7
⇒21x=7
⇒x=\(\dfrac{7}{21}\)
⇒x= \(\dfrac{1}{3}\)
b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
⇒15x-63x2-15+63x + 63x2-35x+36x-20=44
⇒79x-35=44
⇒79x=44+35
⇒79x=79
⇒x=1
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
b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
Mấy cái này chuyển vế đổi dấu là xong í mà :3
1,
16-8x=0
=>16=8x
=>x=16/8=2
2,
7x+14=0
=>7x=-14
=>x=-2
3,
5-2x=0
=>5=2x
=>x=5/2
Mk làm 3 cau làm mẫu thôi
Lúc đăng đừng đăng như v :>
chi ra khỏi ngt nản
từ câu 1 đến câu 8 cs thể làm rất dễ,bn tham khảo bài của bn muwaa r làm những câu cn lại
\(a,\) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+6\)
\(< =>20x-5x^2+5x^2-12-x-6=0\)
\(< =>19x-18=0\)
\(< =>x=\dfrac{18}{19}\)
\(b,\left(2x-7\right)\left(5+4x\right)-8\left(x^2-4x+5\right)=-30\)
\(< =>10x+8x^2-35-28x-8x^2+24x-40+30=0\)
\(< =>6x-45=0< =>x=\dfrac{45}{6}=7,5\)
a) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+\Rightarrow6\\ \Leftrightarrow20x-5x^2+5x^2-12=x+6\\ \Leftrightarrow20x-12=x+6\\\Rightarrow20x-x=6+12\\ \Rightarrow19x=18\\ \Rightarrow x=\dfrac{18}{19}\)
b) \(\left(2x-7\right)\left(5+4x\right)-8\left(x^2-3x+5\right)=-30\\ \Rightarrow10x+8x^2-35-28x-8x^2+24x-40=-30\\ \Rightarrow6x-75=-30\\ \Rightarrow6x=45\\ \Rightarrow x=\dfrac{15}{2}\)
a: \(\Leftrightarrow x\cdot\dfrac{2}{7}-\dfrac{3}{5}x=-1+\dfrac{1}{3}\)
\(\Leftrightarrow x\cdot\dfrac{-11}{35}=\dfrac{-2}{3}\)
hay x=70/33
b: \(\left|3x+7\right|-3x=7\)
\(\Leftrightarrow\left|3x+7\right|=3x+7\)
\(\Leftrightarrow3x+7\ge0\)
hay x>=-7/3