giải pt và bất pt
a) |x+5|=3x+1
b)\(\frac{3\left(x-1\right)}{4}+1\ge\frac{x+2}{3}\)
c)\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
Những câu hỏi liên quan
Giải các PT sau :
a,frac{1-6x}{x-2}+frac{9x+4}{x+2}frac{xleft(3x-2right)+1}{left(x+2right)left(x-2right)}
b,(2 - 3x) (x + 11) (3x - 2) (2 - 5x)
c,frac{3x-2}{6}-5frac{3-2left(x+7right)}{4}
d,frac{x}{x-3}+frac{x}{x
+1}frac{2x}{left(x+1right)left(x-3right)}
e,frac{x+1}{5}+frac{x+2}{4}frac{x+3}{3}+frac{x+4}{2}
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Giải các PT sau :
a,\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{\left(x+2\right)\left(x-2\right)}\)
b,(2 - 3x) (x + 11) = (3x - 2) (2 - 5x)
c,\(\frac{3x-2}{6}-5=\frac{3-2\left(x+7\right)}{4}\)
d,\(\frac{x}{x-3}+\frac{x}{x +1}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
e,\(\frac{x+1}{5}+\frac{x+2}{4}=\frac{x+3}{3}+\frac{x+4}{2}\)
B1 :Giải phương trìnha,frac{3left(x-3right)}{4}-1frac{2x+3left(x+1right)}{6}-frac{7+12x}{12}b,left(x+2right)left(3-4xright)x^2+4x+4c,frac{x-2}{x+2}-frac{3}{x-2}frac{2left(x-11right)}{x^2-4}d,I7-xI-5x1B2:Giải bất phương trìnha,left(x-2right)left(x+2right)ge xleft(x-4right)b,frac{x-1}{4}-1gefrac{x+1}{3}+8
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B1 :Giải phương trình
a,\(\frac{3\left(x-3\right)}{4}-1=\frac{2x+3\left(x+1\right)}{6}-\frac{7+12x}{12}\)
b,\(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
c,\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
d,I7-xI-5x=1
B2:Giải bất phương trình
a,\(\left(x-2\right)\left(x+2\right)\ge x\left(x-4\right)\)
b,\(\frac{x-1}{4}-1\ge\frac{x+1}{3}+8\)
1,Giải PT sau
a,\(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)
b,(x-3)-\(\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
c,\(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\) \(\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
Bài 1:
a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)
\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)
\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)
Suy ra: \(12x-45-12x^2+45x=0\)
\(\Leftrightarrow-12x^2+57x-45=0\)
\(\Leftrightarrow-12x^2+12x+45x-45=0\)
\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)
\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)
mà \(-3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)
b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)
\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)
Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)
\(\Leftrightarrow-x^2+16x-39=0\)
\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)
\(\Leftrightarrow x^2-13x-3x+39=0\)
\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)
Vậy: Tập nghiệm S={3;13}
c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)
\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)
\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)
Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)
\(\Leftrightarrow-21x^2+26x+11=0\)
\(\Leftrightarrow-21x^2-7x+33x+11=0\)
\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)
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2 tháng 7 2020 lúc 17:37
ĐẠI SỐ 1. Giải các phương trình sau :a) frac{25x-655}{95}-frac{5left(x-12right)}{209}frac{89-3x-frac{2left(x-18right)}{5}}{11}b) frac{8left(x+22right)}{45}-frac{7x+149+frac{6left(x+12right)}{5}}{9}frac{x+35+frac{2left(x+50right)}{9}}{5}c) frac{x+frac{2left(3-xright)}{5}}{14}-frac{5x-4left(x-1right)}{24}frac{7x+2+frac{9-3x}{5}}{12}+frac{2}{3}2. Giải các bất phương trình sau :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}+5c) frac{left(3x-2right)^2}{3}-fr...
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ĐẠI SỐ
1. Giải các phương trình sau :
a) \(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
b) \(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
c) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
2. Giải các bất phương trình sau :
a) \(5+\frac{x+4}{5}< x-\frac{x-2}{2}+\frac{x+3}{3}\)
b) \(x+1-\frac{x-1}{3}< \frac{2x+3}{2}+\frac{x}{3}+5\)
c) \(\frac{\left(3x-2\right)^2}{3}-\frac{\left(2x+1\right)^2}{3}\le x\left(x+1\right)\)
d) \(\frac{2x+3}{4}-\frac{x+1}{3}\ge\frac{1}{2}-\frac{3-x}{5}\)
\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)
\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)
\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)
\(< =>3072-107x=\frac{38x-684}{5}\)
\(< =>\left(3072-107x\right)5=38x-684\)
\(< =>15360-535x-38x-684=0\)
\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)
nghệm xấu thế
\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)
\(< =>993-33x-11x-415=0\)
\(< =>578=44x< =>x=\frac{289}{22}\)
Bài 1:
b) Phương trình đã cho tương đương với phương trình:
\(\frac{8\left(x+22\right)-55\left(7x+149\right)-6\left(x+12\right)}{45}=\frac{9\left(x+35\right)+2\left(x+50\right)}{45}\)
\(\Leftrightarrow44x=-1056\)
\(\Leftrightarrow x=-24\)
Vậy x=-24 là nghiệm của phương trình
c) Phương trình đã cho tương đương với phương trình:
\(\frac{3x+6}{70}-\frac{x+4}{24}=\frac{32x+19}{60}+\frac{2}{3}\)
\(\Leftrightarrow12\left(3x+6\right)-35\left(x+4\right)=14\left(32x+19\right)+560\)
\(\Leftrightarrow-447x=894\)
\(\Leftrightarrow x=-2\)
Vậy x=-2 là nghiệm của phương trình
Xem thêm câu trả lời
bài 1:giải các pt sau:
a/frac{1-x}{x+1}+3frac{2x+3}{x+1}
b/frac{left(x+2right)^2}{2x-3}-1frac{x^2+10}{2x-3}
c/frac{1}{x+1}-frac{5}{x-2}frac{15}{left(x+1right)left(2-xright)}
d/frac{1-6x}{x-2}+frac{9x+4}{x+2}frac{xleft(3x-2right)+1}{x^2-4}
e/frac{12}{1-9x^2}frac{1-3x}{1+3x}-frac{1+3x}{1-3x}
ffrac{x+4}{x^2-3x+2}+frac{x+1}{x^2-4x+3}frac{2x+5}{x^2-4x+3}
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bài 1:giải các pt sau:
a/\(\frac{1-x}{x+1}\)+3=\(\frac{2x+3}{x+1}\)
b/\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3}\)
c/\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
d/\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
e/\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
f\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
1,Giải Pt
a,\(\frac{3x-7}{2}+\frac{x+1}{3}=-16\)
b,\(x-\frac{x+1}{3}=\frac{2x+1}{5}\)
c,\(\frac{7-3x}{12}+\frac{3}{4}=2\left(x-2\right)+\frac{5\left(5-2x\right)}{6}\)
e,\(\frac{3\left(x+3\right)}{4}+\frac{1}{2}=\frac{5x+9}{3}-\frac{7x-9}{4}\)
Giải bất phương trình và phương trình sau :
a, \(\left(5x-\frac{2}{3}\right)-\frac{2x^2-x}{2}\ge\frac{x\left(1-3x\right)}{3}-\frac{5x}{4}\)
b, \(\frac{x^2-4-\left|x-2\right|}{2}=x\left(x-1\right)\)
Cho x,y,z là các sô dương.Chứng minh rằng x/2x+y+z+y/2y+z+x+z/2z+x+y<=3/4
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Giải bất phương trình và phương trình sau :
\(a,\left(5x-\frac{2}{3}\right)-\frac{2x^2-x}{2}\ge\frac{x\left(1-3x\right)}{3}-\frac{5x}{4}\)
\(b,\frac{x^2-4-\left|x-2\right|}{2}=x\left(x+1\right)\)
1. Giải PT sau
a) \(\left(\frac{x-1}{x+1}\right)^2-4\left(\frac{x^2-1}{x^2-4}\right)+3\left(\frac{x+1}{x-2}\right)^2=0\)
b) \(\frac{x^2}{3}+\frac{48}{x^2}=10\left(\frac{x}{3}-\frac{4}{x}\right)\)