giai phuong trinh sau: \(\frac{x+2}{5}-\frac{x}{3}=\frac{2x-5}{2}\)
giai bat phuong trinh
a) \(5+\frac{x+4}{5}< x-\frac{x-2}{2}+\frac{x+3}{3}\)
b) \(x+1-\frac{x-1}{3}< \frac{2x+3}{2}+\frac{x}{3}+5\)
giai bat phuong trinh
c) \(\frac{(3x-2)^2}{3}-\frac{\left(2x+1\right)^2}{3}\le x\left(x+1\right)\)
d) \(\frac{2x-3}{4}-\frac{x+1}{3}\ge\frac{1}{2}-\frac{3-x}{5}\)
giai he phuong trinh sau:\(\hept{\begin{cases}\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}\\2\left(x-3\right)-3\left(y+2\right)=-16\end{cases}}\)
\(\hept{\begin{cases}\frac{2x-3y}{2y-5}=\frac{3x+1}{3y-4}\left(1\right)\\2\left(x-3\right)-3\left(y+2\right)=-16\left(2\right)\end{cases}}\)
Nhân chéo và chuyển vế phương trình (1) và nhân phân phối, chuyển vế phương trình (2), ta được:
\(\hept{\begin{cases}7x-11y=-17\\2x-3y=-4\end{cases}}\)
<=>\(\hept{\begin{cases}x=7\\y=6\end{cases}}\)
\(\frac{x}{2x-6}+\frac{x}{2x+2}=\frac{2x^2}{x^2+2x-3}\) giai phuong trinh lop 8
\(\frac{x}{2x-6}+\frac{x}{2x+2}=\frac{2x^2}{x^2+2x-3}\)
\(ĐKXĐ:x^2+2x-3=\left(x+1\right)\left(x-3\right)\\ \Rightarrow x\ne-1;x\ne3\)
\(\frac{x}{2x-6}+\frac{x}{2x+2}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)=4x^2\)
\(\Leftrightarrow x^2+x+x^2-3x=4x^2\)
\(\Leftrightarrow2x^2-2x=4x^2\)
\(\Leftrightarrow2x^2-4x^2-2x=0\)
\(\Leftrightarrow-2x^2-2x=0\)
\(\Leftrightarrow2x\left(-x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=0\\-x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\left(N\right)\\x=-1\left(L\right)\end{cases}}\)
Tự kết luận tập nghiệm bạn nhé!
x2+2x-3 = (x+1)(x-3)
vậy MSC = 2(X+1(X-3) qui đồng mẫu số r làm dc r, đk x khác 1; -3
Giai phuong trinh sau: \(\frac{x^2+2x+2}{x+1}+\frac{x^2+8x+20}{x+4}=\frac{x^2+4x+6}{x+2}+\frac{x^2+6x+12}{x+3}\)
giai phuong trinh
\(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\)
\(ĐKXĐ:x\ne2;x\ne4\)
\(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\)
\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)-\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=\frac{16}{5}\)
\(\Rightarrow\frac{x^2-7x+12-x^2+4x-4}{x^2-6x+8}=\frac{16}{5}\)
\(\Rightarrow\frac{-3x+8}{x^2-6x+8}=\frac{16}{5}\)
\(\Rightarrow-3x+8=\frac{16}{5}\left(x^2-6x+8\right)\)
\(\Rightarrow-3x+8=\frac{16}{5}x^2-\frac{96}{5}x+\frac{128}{5}\)
\(\Rightarrow\frac{16}{5}x^2-\frac{81}{5}x+\frac{88}{5}=0\)
Ta có \(\Delta=\frac{81^2}{5^2}-4.\frac{16}{5}.\frac{88}{5}=\frac{929}{25},\sqrt{\Delta}=\frac{\sqrt{929}}{5}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{81+\sqrt{929}}{32}\\x=\frac{81-\sqrt{929}}{32}\end{cases}}\)
Giai bat phuong trinh
a, \(\frac{2x-5}{3}-\frac{3x-1}{2}< \frac{3-x}{5}-\frac{2x-1}{4}\)
b, \(5x-\frac{3-2x}{2}>\frac{7x-5}{2}+x\)
c, \(\frac{7x-2}{3}-2x< 5-\frac{x-2}{4}\)
d, \(\frac{x}{8}-\frac{x}{4}+\frac{x}{2}>x+5\)
Các bạn ơi giúp mik với đi
mai kiem tra rồi
a,\(\frac{2x-5}{3}-\frac{3x-1}{2}< \frac{3-x}{5}-\frac{2x-1}{4}\)
\(\Leftrightarrow\frac{\left(2x-5\right)20}{60}-\frac{\left(3x-1\right)30}{60}< \frac{\left(3-x\right)12}{60}-\frac{\left(2x-1\right)15}{60}\)
\(\Leftrightarrow40x-100-90x+30< 36-12x-30x+15\)
\(\Leftrightarrow40x-90x+12x+30x< 36+15+100-30\)
\(\Leftrightarrow-8x< 121\)
\(\Leftrightarrow x>-\frac{378}{25}\)
Giai he phuong trinh: \(\hept{\begin{cases}\frac{2}{x}+\frac{5}{x+y}=2\\\frac{3}{x}+\frac{1}{x+y}=1,7\end{cases}}\)
\(\hept{\begin{cases}2.\frac{1}{x}+5.\frac{1}{x+y}=2\\3.\frac{1}{x}+\frac{1}{x+y}=1,7\end{cases}}\)
Đặt \(\frac{1}{x}\)=a
\(\frac{1}{x+y}=b\)
ta có \(\hept{\begin{cases}2a+5b=2\\3a+b=1,7\end{cases}}\)
\(\hept{\begin{cases}a=\frac{1}{2}\\b=\frac{1}{5}\end{cases}}\)
=> \(\frac{1}{x}=\frac{1}{2}\Rightarrow x=2\)
\(\frac{1}{x+y}=\frac{1}{5}\)\(\Rightarrow x+y=5\)\(\Rightarrow y=3\)
giai phuong trinh \(\frac{x+2}{x-5}+3=\frac{6}{2-x}\)