BT2: Tìm x, biết
3) \(\dfrac{1}{2}.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{3}.\left(x-\dfrac{1}{4}\right)\)
4) \(\text{ | }x-\dfrac{1}{3}\text{\text{ |}}-\dfrac{1}{3}=\dfrac{1}{3}\)
\(\text{Tìm x, biết:}\)
\(a\)) \(20\text{%}x-x+\dfrac{1}{5}=\dfrac{3}{4}\)
\(b\)) \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)
\(c\)) \(\left(x-\dfrac{3}{4}\right)\left(4+3x\right)=0\)
\(d\)) \(x-\dfrac{1}{3}x+\dfrac{1}{5}x=\dfrac{-26}{5}\)
\(e\)) \(50\text{%}x+\dfrac{2}{3}x=x-5\)
\(g\)) \(\dfrac{2}{3}\left(x+\dfrac{9}{5}\right)-\dfrac{3}{10}.\left(5x-\dfrac{1}{3}\right)=\dfrac{7}{15}\)
câu c) mang tính mua vui hay gì hả bn
mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)
1) giải phương trình:
a. \(\dfrac{1}{x-5}-\dfrac{3}{x^2-6\text{x}+5}=\dfrac{5}{x-1}\)
b. \(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)
c.\(\dfrac{x}{\left(2\text{x}-3\right)}+\dfrac{x}{2\text{x}+2}=\dfrac{2\text{x}}{\left(x+1\right)\left(x-3\right)}\)
d.\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5\text{x}-2}{4-x^2}\)
e. \(\dfrac{1-6\text{x}}{x-2}+\dfrac{9\text{x}+4}{x+2}=\dfrac{x\left(3\text{x}-2\right)+1}{x^2-4}\)
rút gọn các biểu thức sau
\(B=\dfrac{3\text{x}^2+6\text{x}+12}{x^3-8\dfrac{ }{ }}\)
C=\(\left(\dfrac{x+1}{2\text{x}-2}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2\text{x}+2}\right).\dfrac{4\text{x}^2-4}{5}\)
E=\(\dfrac{x^2-10\text{x}+25}{x^2-5\text{x}}\)
c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
tính giá trị của biểu thức
a) \(A=2x^2-\dfrac{1}{3}y,t\text{ại}x=2;y=9\)
b) \(P=2x^2+3xy+y^2t\text{ại }x=-\dfrac{1}{2};y=\dfrac{2}{3}\)
c) \(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)t\text{ại}x=2;y=\dfrac{1}{4}\)
a) \(A=2x^2-\dfrac{1}{3}y\)
A= \(\left(2-\dfrac{1}{3}\right)\)\(x^2y\)
A=\(\dfrac{5}{3}\)\(x^2y\)
Tại \(x=2;y=9\) ta có
A=\(\dfrac{5}{3}\).(2)\(^2\).9 = \(\dfrac{5}{3}\).4 .9 = 60
Vậy tại \(x=2;y=9\) biểu thức A= 60
b) P=\(2x^2+3xy+y^2\) (\(y^2\) là 1\(y^2\) nha bạn)
P=\(\left(2+3+1\right)\left(x^2.x\right)\left(y.y^2\right)\)
P= 6\(x^3y^3\)
Tại \(x=-\dfrac{1}{2};y=\dfrac{2}{3}\) ta có
P= 6.\(\left(-\dfrac{1}{2}\right)^3.\left(\dfrac{2}{3}\right)^3\) = 6.\(\left(-\dfrac{1}{8}\right).\dfrac{8}{27}\) = \(-\dfrac{2}{9}\)
Vậy tại \(x=-\dfrac{1}{2};y=\dfrac{2}{3}\) biểu thức P= \(-\dfrac{2}{9}\)
c)\(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)\)
=\(\left((-\dfrac{1}{2}).\dfrac{2}{3}\right)\left(x.x^3\right).y^2\)
=\(-\dfrac{1}{3}\)\(x^4y^2\)
Tại \(x=2;y=\dfrac{1}{4}\)ta có
\(-\dfrac{1}{3}\).\(\left(2\right)^4.\left(\dfrac{1}{4}\right)^2=-\dfrac{1}{3}.16.\dfrac{1}{16}=-\dfrac{1}{3}\)
\(\)Vậy \(x=2;y=\dfrac{1}{4}\) biểu thức \(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)\)= \(-\dfrac{1}{3}\)
CHÚC BẠN HỌC TỐT NHA
Giải các phương trình sau:
1. \(x-\dfrac{\dfrac{x}{2}-\dfrac{3+x}{4}}{2}=\dfrac{2\text{x}-\dfrac{10-7\text{x}}{3}}{2}-x-1\)
2.\(\dfrac{3}{10}\left(1,2-x\right)-\dfrac{5+7x7}{4}=\dfrac{1}{20}\left(9\text{ }x+0,2\right)-\dfrac{12,5x+4,5}{5}\)
1 a..Rút gọn biểu thức A = \(\dfrac{\text{ x 2 − 4 x + 4}}{\text{x 3 − 2 x 2 − ( 4 x − 8 ) }}\)
b. Rút gọn biểu thức B = \(\left(\dfrac{x+2}{\text{x }\sqrt{\text{x }}+1}-\dfrac{1}{\sqrt{\text{x}}+1}\right).\dfrac{\text{4 }\sqrt{x}}{3}\)
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
Tìm x:
\(a\)) \(\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)
\(b\)) \(\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{27}{8}\right)^3=\dfrac{81}{16}\)
\(c\)) \(\dfrac{1}{2}.2^x+4.2^x=9.2^5\)
\(d\)) \(\text{12 - (2x +1)}^2=-69\)
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)
\(x=\dfrac{1}{2}+\dfrac{1}{3}\)
\(x=\dfrac{1}{5}\)
\(\)\(\text{Tìm x:}\)
a) \(x+\dfrac{1}{2}=2\)
b) \(75\%x-\dfrac{1}{2}=-1\dfrac{1}{4}\)
c) \(\left|x-\dfrac{2}{5}\right|+\dfrac{7}{10}=3\)
Cho biểu thức:\(P=\left(\dfrac{2\text{x}}{x^2-9}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x^2-3\text{x}}\right)v\text{ới}x\ne\pm3;x\ne0;x\ne5\)
1, Chứng minh \(P=\dfrac{x}{x-5}\)
1: \(P=\left(\dfrac{2x}{x^2-9}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x^2-3x}\right)\)
\(=\left(\dfrac{2x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x\cdot\left(x-3\right)}\right)\)
\(=\dfrac{2x-x+3}{\left(x-3\right)\left(x+3\right)}:\dfrac{2\left(x-3\right)-x+1}{x\left(x-3\right)}\)
\(=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(x-3\right)}{2x-6-x+1}\)
\(=\dfrac{x}{x-5}\)