(2x-6)(x2+2)=(2x-6)(8x-10)
(5x-1)2=(3x+5)2
109-x/91 +107-x/93 +105-x/95+103-x/97=-4
1)Tìm GTNN của A = 5x^2 + 5y^2 + 6x - 6y - 2xy
2 )\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}=-4\)Tìm x
(109-x)/91+(107-x)/93+(105-x)/95+(103-x)/97=-4
[(109-x)/91 +1]+[(107-x)/93 +1]+[(105-x)/95 +1]+[(103-x)/97 +1]-4=-4
(109+91-x)/91+(107+93-x)/93+(105+95-x)/95+(103+97-x)/97=-4+4
(200-x)/91+(200-x)/93+(200-x)/95+(200-x)/97=0
(200-x)(1/91+1/93+1/95+1/97)=0
Ma : 1/91+1/93+1/95+1/97\(\ne\)0
=>200-x=0
=>x=200
Giải phương trình sau
a,\(2\left(\frac{11x}{12}+\frac{1}{3}\right)=2-\frac{x}{6}\)
b,\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
c,\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}+4=0\)
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, \(2\left(\frac{11x}{12}+\frac{1}{3}\right)=2-\frac{x}{6}\)
\(2\left(\frac{11x+4}{12}\right)-2+\frac{x}{6}=0\)
\(\frac{44x+8}{12}-2+\frac{x}{6}=0\)
\(\frac{44x+8}{12}-\frac{24}{12}+\frac{2x}{12}=0\)
\(\frac{44x+8-24+2x}{12}=\frac{46x-16}{12}=0\)
\(\Leftrightarrow46x-16=0\)
\(\Leftrightarrow46x=16\Rightarrow x=\frac{8}{23}\)
Vậy nghiệm của pt là x=8/23
k mk
Giải phương trình sau
\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}+4=0\)
\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{97}+\frac{x+100}{96}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
Dễ thấy \(\left(\frac{1}{99}< \frac{1}{98}< \frac{1}{97}< \frac{1}{96}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)\ne0\)
\(\Rightarrow x+100=0\Rightarrow x=-100\)
Vậy x = -100
\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}+4=0\)
\(\Rightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)
\(\Rightarrow\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)
\(\Rightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)=0\)
Dễ thấy \(\left(\frac{1}{91}>\frac{1}{93}>\frac{1}{95}>\frac{1}{97}\right)\)nên \(\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)\ne0\)
\(\Rightarrow200-x=0\Rightarrow x=200\)
Vậy x = 200
Giải Phương trình
1)\(\dfrac{99-x}{101}+\dfrac{97-x}{103}+\dfrac{95-x}{105}+\dfrac{93-x}{107}=4\)
2)\(\dfrac{x+14}{86}+\dfrac{x+15}{85}+\dfrac{x+16}{84}+\dfrac{x+17}{83}+\dfrac{x+116}{4}=0\)
3)\(x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x-\dfrac{1-2x}{3}}{5}\)
Bài 1:
\(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)
\(\Leftrightarrow \frac{99-x}{101}+1+\frac{97-x}{103}+1+\frac{95-x}{105}+1+\frac{93-x}{107}+1=0\)
\(\Leftrightarrow \frac{99-x+101}{101}+\frac{97-x+103}{103}+\frac{95-x+105}{105}+\frac{93-x+107}{107}=0\)
\(\Leftrightarrow \frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)
\(\Leftrightarrow (200-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\neq 0\) nên suy ra \(200-x=0\Rightarrow x=200\)
Bài 2:
\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+116}{4}=0\)
\(\Leftrightarrow \frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow \frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow (x+100)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
Vì \(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\neq 0\). Do đó \(x+100=0\Rightarrow x=-100\)
1/
99-x/101+97-x/103+95-x/105+93-x/107
\(\Leftrightarrow\)(99-x/101+1)+(97-x/103+1)+(95-x/105+1)+(93-x/107+1)=0
\(\Leftrightarrow\)100-x/101+100-x/103+100-x/105+100-x/107=0
\(\Leftrightarrow\)(100-x)(1/101+1/103+1/105+1/107)=0
\(\Leftrightarrow\)100-x=0\(\Leftrightarrow\)
x=-100
Cho A=\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}\)và B=-2
Tìm x để A=B
\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}=-2\)
\(\Leftrightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=2\)
\(\Leftrightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=2\)
\(\Rightarrow x=200-\frac{2}{\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}}\)
Bạn tự bấm máy, kết quả chẳng đẹp gì
\(A=2B\) thì còn có lý
(99-x)/101+(97-x)/103+(95-x)/105+(93-x)/107=-4
Tim X
3) -12 + (2x – 9) + x= 0
4) 11 + (15 - x) = 1
5) 4 - (27 - 3) = x - (13 - 4)
6) 8 - (x - 10) = 23 - (- 4 +12)
7) 105 – 5(10 – 5x) = -20
8) (x -1)(8-2x)(3x+123) = 0
9) (x2 - 25)(x+ 10) = 0
10) x(x2+5) =
3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)
4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)
5)
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
\(3.-12+\left(2x-9\right)+x=0.\)
\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)
Vậy \(x=7.\)
\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)
Vậy \(x=25.\)
\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)
Vậy \(x=-11.\)
\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)
Vậy \(x=3.\)
\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)
Vậy \(x=-3.\)
\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)
Vậy \(x\in\left\{1;4;-41\right\}.\)
\(9.\left(x^2-25\right)\left(x+10\right)=0.\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)
Vậy \(x\in\left\{5;-5;-10\right\}.\)
\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)
1.(x+2)3+(x-3)2-x2(x+5)
2.(2x+3).(x-5)+2x(3-x)+x-10
3.(x+5).(x2-5x+25)-x(x-4)2+16x
4.(-x-2)3+(2x-4).(x2+2x+4)-x2.(x-6)
3: \(\left(x+5\right)\left(x^2-5x+25\right)-x\left(x-4\right)^2+16x\)
\(=x^3+125-x^3+8x^2-16x+16x\)
\(=8x^2+125\)
giải các phương trình sau:
a) 4x.(2x+3) - x.(8x-1)= 5.(x+2)
b) (3x-5).(3x+5) - x.(9x-1) = 4
c) 3 - 4x . (25-2x) = 8x^2 + x - 300
d) 2. (1-3x/5) - 2+3x/10 = 7-3.(2x+1)/4
e) 5x + 2/6 - 8x - 1 /3 = 4x + 2 /5 - 5
f) 5x - 4/2 = 16x + 1/7
g) 12x + 5 / 3=2x - 7/4
h) 3t - 8/12 = 5-t /8
i) 5u + 6/15 = u - 4 /10
k) 3.(x-11)/4 = 3. (x+1) - 2.(2x+ 5)/10
l) 14 và 1/2 - 2. (x+3)/5 = 3x/2 - 2. (x-7 )/3
m ) 2x -5 / 6 - x +2 = 5x - 3/ 3 - 6x - 7/ 4 + x
n) x-4/5 + 3x - 2/10 - x = 2x-5/3 - 7x+2/6
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