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Ta có\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+9}{91}=\frac{x+91}{9}+\frac{x+92}{8}+\frac{x+61}{39}\)
<=> \(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+9}{91}+1\right)=\left(\frac{x+91}{9}+1\right)+\left(\frac{x+92}{8}+1\right)+\left(\frac{x+61}{39}+1\right)\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}=\frac{x+100}{9}+\frac{x+100}{8}+\frac{x+100}{39}\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}-\frac{x+100}{9}-\frac{x+100}{8}-\frac{x+100}{39}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\right)=0\)
Do \(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\ne0\)
Nên x+100=0 => x=-100
4.
\(\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+5}{95}=\dfrac{x+7}{93}+\dfrac{x+9}{91}+\dfrac{x+11}{89}\\ \Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+5}{95}+1\right)=\left(\dfrac{x+7}{93}+1\right)+\left(\dfrac{x+9}{91}+1\right)+\left(\dfrac{x+11}{89}+1\right)\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{97}++\dfrac{x+100}{95}=\dfrac{x+100}{93}+\dfrac{x+100}{91}+\dfrac{x+100}{89}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{93}-\dfrac{1}{91}-\dfrac{1}{89}\right)=0\\ \Leftrightarrow x+100=0\Leftrightarrow x=-100\)
\(\text{1) }\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\\ \Leftrightarrow\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\cdot24=\left[\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\right]24\\ \Leftrightarrow3\left(4x^2-9\right)=4\left(x^2-8x+16\right)+8\left(x^2-4x+4\right)\\ \Leftrightarrow12x^2-27=4x^2-32x+64+8x^2-32x+32\\ \Leftrightarrow12x^2-27=12x^2-64x+96\\ \Leftrightarrow12x^2-12x^2+64x=96+27\\ \Leftrightarrow64x=123\\ \Leftrightarrow x=\dfrac{123}{64}\\ \text{Vậy }S=\left\{\dfrac{123}{64}\right\}\\ \)
\(\text{2) }x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}\\ \Leftrightarrow\left(x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}\right)15=\dfrac{7x-\dfrac{x-3}{2}}{5}\cdot15\\ \Leftrightarrow15x+30-2x-\dfrac{2x-5}{6}=21x-\dfrac{3x-9}{2}\\ \Leftrightarrow15x-2x-\dfrac{2x-5}{6}-21x+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow\left(-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}\right)6=-30\cdot6\\ \Leftrightarrow-48x-2x+5+9x-27=-180\\ \Leftrightarrow-41x==-158\\ \Leftrightarrow x=\dfrac{158}{41}\\ \text{Vậy }S=\left\{\dfrac{158}{41}\right\}\)
\(\text{3) }1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7}{3}}{2}\\ \Leftrightarrow\left(1-\dfrac{x-1-x}{3}\right)6=\left(\dfrac{x}{2}-\dfrac{2x-1}{2}\right)6\\ \Leftrightarrow6+2=-3x+3\\ \Leftrightarrow-3x=8-3\\ \Leftrightarrow-3x=5\\ \Leftrightarrow x=-\dfrac{5}{3}\\ \\ \text{Vậy }S=\left\{-\dfrac{5}{3}\right\}\)
a.
\(\dfrac{5x-17}{14}+\dfrac{x-3}{26}>\dfrac{29-9x}{91}\)
\(\Leftrightarrow13\left(5x-17\right)+7\left(x-3\right)>2\left(29-9x\right)\)
\(\Leftrightarrow65x-221+7x-21>58-18x\)
\(\Leftrightarrow65x+7x+18x>58+21+221\)
\(\Leftrightarrow90x>300\)
\(\Leftrightarrow x>\dfrac{10}{3}\)
b)
\(\dfrac{8x-1}{9}+\dfrac{3x-2}{4}< \dfrac{43+8x}{12}+\dfrac{35x}{36}\)
\(\Leftrightarrow4\left(8x-1\right)+9\left(3x-2\right)< 3\left(43+8x\right)+35x\)
\(\Leftrightarrow32x-4+27x-18< 129+24x+35x\)
\(\Leftrightarrow32x+27x-24x-35x< 129+18+4\)
\(\Leftrightarrow0x< 151\) ( luôn đúng)
Vậy bất pt vô số nghiệm
Bài1:
\(a,\left(-8\right)^9\) và \(\left(-32\right)^5\)
Ta có:
\(\left(-8\right)^9=-2^{27}\)
\(\left(-32\right)^5=\left(-8.4\right)^5=-2^{27}.2^{10}\)
Vì \(-2^{27}.10< -2^{27}\) nên \(\left(-8\right)^9>\left(-32\right)^5\)
Các câu sau tương tự
Bài2:
\(a,2\left|x-1\right|-3x=7\)
+)Xét \(x\ge1\Rightarrow\left|x-1\right|=x-1\)
Do đó:
\(2\left(x-1\right)-3x=7\\ \Leftrightarrow2x-2-3x=7\\ \Leftrightarrow-x=9\\ \Leftrightarrow x=-9\left(loại\right)\)
+)Xét \(x< 1\Rightarrow\left|x-1\right|=1-x\)
Do đó:
\(2\left(1-x\right)-3x=7\\ \Leftrightarrow2-2x-3x=7\\ \Leftrightarrow-5x=5\\ x=-1\left(chon\right)\)
Vậy x=-1
Câu b tương tự
Bài 1:
\(a,\left(-8\right)^9\) và \(\left(-32\right)^5\)
\(\left(-8\right)^9=\left[\left(-2\right)^3\right]^9=\left(-2\right)^{27}\)
\(\left(-32\right)^5=\left[\left(-2\right)^5\right]^5=\left(-2\right)^{25}\)
\(\left(-2\right)^{27}< \left(-2\right)^{25}\)
\(\Rightarrow\left(-8\right)^9< \left(-32\right)^5\)
\(b,2^{21}\) và \(3^{14}\)
\(2^{21}=\left(2^3\right)^7\)
\(3^{14}=\left(3^2\right)^7\)
\(2^3< 3^2\)\(\Rightarrow2^{21}< 3^{14}\)
\(c,12^8\) và \(8^{12}\)
\(12^8=\left(12^2\right)^4=144^4\)
\(8^{12}=\left(8^3\right)^4=512^4\)
\(144^4< 512^4\)\(\Rightarrow12^8< 8^{12}\)
\(d,\left(-5\right)^{39}\) và \(\left(-2\right)^{91}\)
\(\left(-5\right)^{39}=\left[\left(-5\right)^3\right]^{13}\)
\(\left(-2\right)^{91}=\left[\left(-2\right)^7\right]^{13}\)
\(\left(-5\right)^3>\left(-2\right)^7\)\(\Rightarrow\left(-5\right)^{39}>\left(-2\right)^{91}\)
Bài 2:
\(a,2.\left|x-1\right|-3x=7\)
\(\left|x-1\right|=\dfrac{7+3x}{2}\)
Ta có 2 trường hợp:
Th1:\(x-1=\dfrac{7-3x}{2}\)
\(\dfrac{2x-2}{2}=\dfrac{7+3x}{2}\)
\(\Rightarrow2x-2=7+3x\)
\(2x-3x=7+2\)
\(-x=9\Rightarrow x=-9\)
Th2:\(x+1=-\dfrac{7+3x}{2}\)
\(\dfrac{2x-2}{2}=\dfrac{-7-3x}{2}\)
\(\Rightarrow2x-2=-7-3x\)
\(2x+3x=-7+2\)
\(5x=-5\Rightarrow x=-1\)
Vậy \(x\in\left\{-9;-1\right\}\)
\(b,\left|5x-3\right|=\left|7-x\right|\)
Ta có: Th1: \(\left|7-x\right|=7-x\) khi \(7-x\ge0\)\(\Rightarrow x\le7\)
\(5x-3=7-x\)
\(5x+x=7+3\)
\(6x=10\Rightarrow x=\dfrac{10}{6}=\dfrac{5}{3}\)( thoả mãn )
vì x thoả mãn \(x\le7\)\(\Rightarrow\) th1 thoả mãn x
Ta có: Th2: \(\left|7-x\right|=-\left(7-x\right)\) khi \(7-x< 0\Rightarrow x>7\)
\(5x-3=-\left(7-x\right)\)
\(5x-3=-7+x\)
\(5x-x=-7+3\)
\(4x=-4\Rightarrow x=-1\) ( loại )
Vì x thoả mãn \(x>7\) mà \(x=-1\Rightarrow\)th2 loại
a) \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3x-24}\) \(ĐK:x\ne8\)
\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3\left(x-8\right)}\)
\(\Leftrightarrow\frac{3.3}{6.\left(x-8\right)}+\frac{6.\left(3x-20\right)}{6\left(x-8\right)}-\frac{2\left(3x-102\right)}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow\frac{9+18x-120-6x+204}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow\frac{12x+93}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow8\left(12x+93\right)=-6\left(x-8\right)\)
\(\Leftrightarrow96x+744=-6x+48\)
\(\Leftrightarrow102x=-696\)
\(\Leftrightarrow x=\frac{-116}{17}\) (nhận)
Vậy .....
b) \(\frac{1}{3-x}+\frac{14}{x^2-9}=\frac{x-4}{3+x}+\frac{7}{3+x}\) \(ĐK:x\ne\pm3\)
\(\Leftrightarrow\frac{1}{3-x}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{x-4}{3+x}+\frac{7}{3+x}\)
\(\Leftrightarrow-\frac{3+x}{\left(x-3\right)\left(3+x\right)}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{-3-x+14}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)
\(\Leftrightarrow-3-x+14=x^2-3x-4x+12+7x-21\)
\(\Leftrightarrow x=-5\) (nhận)
Vậy ....
......................?
mik ko biết
mong bn thông cảm
nha ................
b, \(\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\frac{x+200}{99}+\frac{x+200}{98}=\frac{x+200}{97}+\frac{x+200}{96}\)
\(\frac{x+200}{99}+\frac{x+200}{98}-\frac{x+200}{97}-\frac{x+200}{96}=0\)
\(\left(x+200\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
mà\(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\ne0\)
==> x+200=0
<=>x=-200
Vậy nghiệm của phương trình là x=-200
c, \(\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)
\(\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)
\(\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
mà \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
==>200-x=0
<=>x=200
vậy nghiệm của pt là x=200
a) ( 100 – 1 ) 3 = 970299. b) ( 91 + 9 ) 3 = 100 3 .
c) ( 1000 + 1 ) 3 = 1003003001. d) ( 102 – 2 ) 3 = 100 3 .