\(\left(3x-1\right).\left(\frac{-1}{2}x+5\right)=0\)
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)
\(\left(2x+\frac{3}{5}\right)^{^2}-\frac{9}{25}=0\)
\(3\left(3x-\frac{1}{2}\right)+\frac{1}{9}=0\)
Tìm x
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
giải nhanh hộ mình với, mai mình nộp rồi
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{3^2}{5^2}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow\hept{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=\frac{3}{5}-\frac{3}{5}\\2x=-\frac{3}{5}-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=0\\2x=\frac{-6}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0:2\\x=-\frac{6}{5}:2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=0-\frac{1}{9}\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}:3\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1^3}{3^3}\right)\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=-\frac{1}{3}\)
\(\Rightarrow3x=-\frac{1}{3}+\frac{1}{2}\)
\(\Rightarrow3x=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{6}:3\)
\(\Rightarrow x=\frac{1}{18}\)
Bài 1:Giải phương trình
a)\(10x^2-5x\left(2x+3\right)=15\)
b)\(3x-7-\left(3-4x\right)\left(2x+1\right)=4x\left(2x-7\right)\)
c)\(\left(4x-5\right)^2-\left(7-2x\right)=4\left(2x-4\right)^2+6x\)
Bài 2:Giải phương trình
a)\(\frac{3\left(x-1\right)}{2}+4=\frac{2x}{3}+\frac{4-5x}{6}\)
b)\(\frac{4-x}{7}-\frac{1}{7}\left(\frac{7+3x}{9}+\frac{5-2x}{2}\right)=4-\frac{4x}{3}\)
c)\(\frac{2}{9}\left(2x-5\right)-\frac{5}{3}\left[\left(x-2\right)-\frac{7}{12}\right]=\frac{3}{4}\left(x-3\right)\)
Bài 3:Giải phương trình
a)\(\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\)
b)\(2x\left(x-3\right)+5\left(x-3\right)=0\)
c)\(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
Bài 4:Tìm m để phương trình sau có nghiệm bằng 7:\(\left(2m-5\right)x-2m^2+8=43\)
Bài 5:Giải phương trình
a)\(\left(2x-1\right)^2-\left(2x+1\right)^2=0\)
b)\(\frac{1}{27}\left(x-3\right)^3-\frac{1}{125}\left(x-5\right)^3=0\)
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
Bài 4 xem lại đề nhé bác
Tìm x:
a)\(2.\left(3x-\frac{1}{2}\right)-2x=\frac{1}{2}\left(2x-3\right)\)
b)\(\left(2x-\frac{3}{5}\right)^2=\frac{4}{25}\)
c)\(\left(3x-1\right)^3=27\)
d)\(5-\left|x\right|=2\)
e)|2x+1|-3=3
f)|3-2x|=5
\(\left(5-x\right)\left(3x-\frac{1}{4}\right)=0\)
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
1) giải phương trình:
a) \(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x+5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
b) \(\frac{7x+10}{x+1}\left(x^2-x-2\right)-\frac{7x+10}{x+1}\left(2x^2-3x-5\right)=0\)
c) \(\frac{2x+5}{x+3}+1=\frac{4}{x^2+2x-3}-\frac{3x-1}{1-x}\)
d) \(\frac{13}{2x^2+x-21}+\frac{1}{2x+7}+\frac{6}{9-x^2}=0\)
e) \(\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)
f) \(\frac{1+\frac{x}{x+3}}{1-\frac{x}{x+3}}=3\)
Bài 1 : Giải bất phương trình sau
1 , \(\left(2x+3\right)\left(5x-7\right)\ge0\)
2 , \(\left(3-2x\right)\left(4x+3\right)< 0\)
3 , \(\left(2x+5\right)\left(3-x\right)\left(5x-1\right)\le0\)
4 , \(x^2-3x+2< 0\)
5 , \(-x^2+12x+13>0\)
6 , \(x^2+6x+9\le0\)
7 , \(\frac{x+2}{3x+1}>\frac{x-2}{2x-1}\)
8 , \(\frac{1}{x+2}< \frac{3}{x-3}\)
9 , \(\frac{5x-6}{2x-5}\le6\)
10 , \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)
Tìm x :
a) \(\frac{3x+2}{2}-\frac{3x+1}{6}=2x+\frac{5}{3}\)
b) \(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
c) \(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
d) \(\left(x+1\right)^2-4\left(x^2-2x+1\right)=0\)
a) Qui đồng rồi khử mẫu ta được:
3(3x+2)-(3x+1)=2x.6+5.2
<=> 9x+6-3x-1 = 12x+10
<=> 9x-3x-12x = 10-6+1
<=> -6x = 5
<=> x = -5/6
Vậy ....
b) ĐKXĐ: \(x\ne\pm2\)
Qui đồng rồi khử mẫu ta được:
(x+1)(x+2)+(x-1)(x-2) = 2(x2+2)
<=> x2+3x+2+x2-3x+2 = 2x2+4
<=> x2+x2-2x2+3x-3x = 4-2-2
<=> 0x = 0
<=> x vô số nghiệm
Vậy x vô số nghiệm với x khác 2 và x khác -2
c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)
\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)
Vậy ......
d) (x+1)2-4(x2-2x+1) = 0
<=> x2+2x+1-4x2+8x-4 = 0
<=> -3x2+10x-3 = 0
giải phương trình
a)4:(X-1)=(X-1):9
b) \(\frac{3^0}{5}+\frac{1}{2}:\left(3-\frac{1}{3}\right)^{\sqrt{9}}+\frac{1}{3}\)
c) \(3\frac{1}{2}:x\cdot\frac{1}{2}=5\frac{1}{3}:\frac{1}{2}\cdot\frac{1}{5}\)
d) /2x-3/=5
e) /2-/3x-1//=5
f)\(\left(2x-\frac{1}{3}\right)^2=\left(1-3x\right)^2\)
g) \(\left(3-\frac{1}{2}:x\right)^2=\frac{1}{4}\)
h) \(\left(1-1\frac{1}{2}x\right)^3=\frac{1}{8}\)
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
từ phần sau số 0 thứ nhất là câu mới nha