\(\frac{1}{2}\left(x-6,2\right)-3,4\left(x+0,7\right)=0\)
Những câu hỏi liên quan
\(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
và
\(\left|x-3,4\right|+\left|2.6-x\right|=0\)
\(\frac{\left(-3\right)-\frac{1}{2}\cdot\left(7\frac{1}{2}-5,3\right)\cdot2-3x}{-6,2+2\cdot\left(\frac{1}{2}+1,6\right)}=10,9+x\)
\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{4x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(x^2-3x=0\)
đâu phải toán lớp 1
bạn chọn nhầm à
Bài 2. Tìm x biết: \(\left\{\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{63.65}\right\}.13-\orbr{ }3,08:\left(x+1,2\right):0,7=\frac{1}{5}\)
22 tháng 2 2022 lúc 16:25
Tìm x, biết:
\(a,\dfrac{1}{3}:\left(2x-1\right)=\dfrac{-1}{6}\)
\(b,\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)
\(c,\dfrac{x}{8}=\dfrac{9}{4}\)
\(d,\dfrac{x-3}{2}=\dfrac{18}{x-3}\)
\(e,4,5x-6,2x=6,12\)
\(h,11,4-\left(x-3,4\right)=-16,2\)
a: =>2x-1=-2
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}\)
c: x/8=9/4
nên x/8=18/8
hay x=18
d: \(\Leftrightarrow\left(x-3\right)^2=36\)
=>x-3=6 hoặc x-3=-6
=>x=9 hoặc x=-3
e: =>-1,7x=6,12
hay x=-3,6
h: =>x-3,4=27,6
hay x=31
Đúng 1
Bình luận (0)
a) \(\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}\)
\(\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=-1\left(2x-1\right)\div6\)
\(\dfrac{1}{3}=-2x+1\div6\)
\(x=-\dfrac{1}{2}\)
b) \(\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)
\(TH1:3x+2=0\)
\(3x=0-2\)
\(3x=-2\)
\(x=\dfrac{-2}{3}\)
\(TH2:\left(-\dfrac{2}{5}x-7\right)=0\)
\(\left(\dfrac{-2}{5}x-7\right)=0\)
\(\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0\)
\(\left(\dfrac{-2x-35}{5}\right)=0\)
\(-2x-35=0\)
\(-2x=0+35\)
\(x=-\dfrac{35}{2}\)
c) \(\dfrac{x}{8}=\dfrac{9}{4}\)
\(\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18\)
\(x=18\)
d) \(\dfrac{x-3}{2}=\dfrac{18}{x-3}\)
\(x-3=18+2\)
\(x=20-3\)
\(x=17\)
e) \(4,5x-6,2x=6,12\)
\(\dfrac{9x}{2}-6,2.x=6,12\)
\(\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12\)
\(\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12\)
\(\dfrac{45x-62x}{10}=6.12\)
\(=-17x\div10=6.12\)
\(-17x=10.6.12\)
\(x=-3,6\)
h) \(11,4-\left(x-3,4\right)=-16,2\)
\(x-3,4=-16,2+11,4\)
\(x-3,4=-4,8\)
\(x=-1,4\)
Đúng 2
Bình luận (0)
bt em gửi cô Thương1)ĐKXĐhept{begin{cases}xne1xne3end{cases}} frac{x+5}{x-1}frac{x+1}{x-3}-frac{8}{x^2-4x+3}Leftrightarrowfrac{x+5}{x-1}-frac{x+1}{x-3}+frac{8}{x^2-4x+3}0Leftrightarrowfrac{x+5}{x-1}-frac{x+1}{x-3}+frac{8}{x^2-x-3x+3}0Leftrightarrowfrac{left(x+5right)left(x-3right)}{left(x-1right)left(x-3right)}-frac{left(x+1right)left(x-1right)}{left(x-3right)left(x-1right)}+frac{8}{xleft(x-1right)-3left(x-1right)}0Leftrightarrowfrac{x^2+2x-15}{left(x-1right)left(x-3right)}-frac{x^2-1}{left(x-3r...
Đọc tiếp
bt em gửi cô Thương
1)\(ĐKXĐ\hept{\begin{cases}x\ne1\\x\ne3\end{cases}}\)
\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)
\(\Leftrightarrow\frac{x+5}{x-1}-\frac{x+1}{x-3}+\frac{8}{x^2-4x+3}=0\)
\(\Leftrightarrow\frac{x+5}{x-1}-\frac{x+1}{x-3}+\frac{8}{x^2-x-3x+3}=0\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}+\frac{8}{x\left(x-1\right)-3\left(x-1\right)}=0\)
\(\Leftrightarrow\frac{x^2+2x-15}{\left(x-1\right)\left(x-3\right)}-\frac{x^2-1}{\left(x-3\right)\left(x-1\right)}+\frac{8}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{2x-6}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\)( tm)
Vậy nghiemj của pt x=3
2)\(x^3-x^2-9x+9=0\)
\(\Leftrightarrow x^2\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)hoặc x+3=0
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)hoặc x=-3
Vậy tập hợp nghiệm \(S=\left\{1;3;-3\right\}\)
Bài 1 dài dòng quá em :( Rút gọn bớt cũng được thì phải
Chị ơi bài 1 em sai cái gì ko ạ ? đk x khác 3 mà đúng ko
Bài 1 em không làm sai gì nhưng kết quả sai. Vì đk # 3 nên kết x = 3 không thỏa mãn em ơi :v
Xem thêm câu trả lời
\(0=-\frac{\left(x+2\right)^2+12}{\left(x+2\right)^2}+\frac{\left(x+1\right)^2+1}{\left(x+1\right)^2}-\frac{\left(x+3\right)^2+3}{\left(x+3\right)^2}+\frac{\left(x+4\right)^2+4}{\left(x+4\right)^2}\)
1/Rút gọn
Afrac{left(sqrt{x}+1right)left(x-sqrt{xy}right)left(sqrt{x}+sqrt{y}right)}{left(x-yright)left(sqrt{x^3+x}right)}(x0; y0; x#y)
B left(frac{1}{sqrt{x}+1}-frac{1}{x+sqrt{x}}right):frac{x-sqrt{x}+1}{xsqrt{x}+1}( x0)
Cleft(frac{x+1}{sqrt{x}}+2right).frac{sqrt{x}}{left(sqrt{x}+1right)left(xsqrt{x}+1right)}(x0)
Dleft(frac{xsqrt{x}-1}{sqrt{x}-1}+sqrt{x}right):left(x-1right)-frac{2}{sqrt{x}-1}(x0; x#1)
giúp em với ạ em đang cần gấp ạ
Đọc tiếp
1/Rút gọn
A=\(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{xy}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\left(\sqrt{x^3+x}\right)}\)(x>0; y>0; x#y)
B= \(\left(\frac{1}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right):\frac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)( x>0)
C=\(\left(\frac{x+1}{\sqrt{x}}+2\right).\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x\sqrt{x}+1\right)}\)(x>0)
D=\(\left(\frac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right):\left(x-1\right)-\frac{2}{\sqrt{x}-1}\)(x>=0; x#1)
giúp em với ạ em đang cần gấp ạ
Tìm x biết
a)\(\frac{x+1}{x-4}>0\)
b)\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
c)\(\left(x+2\right)\left(x-3\right)< 0\)
d)\(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{5}\right|+\left|z+\frac{1}{2}\right|\le0\)
Ta có : \(\frac{x+1}{x-4}>0\)
Thì sảy ra 2 trường hợp
Th1 : x + 1 > 0 và x - 4 > 0 => x > -1 ; x > 4
Vậy x > 4
Th2 : x + 1 < 0 và x - 4 < 0 => x < -1 ; x < 4
Vậy x < (-1) .
Đúng 0
Bình luận (0)
Ta có : \(\left(x+2\right)\left(x-3\right)< 0\)
Th1 : \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\left(\text{Vô lý }\right)}\)
Th2 : \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3}\)
Đúng 0
Bình luận (0)
\(\Rightarrow\frac{x-4}{x-4}+\frac{5}{x-4}>0\)
\(\Rightarrow1+\frac{5}{x-4}>0\)
\(\Rightarrow\frac{5}{x-4}>-1\)
\(\Rightarrow\frac{-5}{-x+4}>-\frac{5}{5}\)
\(\Rightarrow-x+4< -5\)
\(\Rightarrow-x< -9\)
\(\Rightarrow x>9\)
Đúng 0
Bình luận (0)