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\(a)5-\left(x-6\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-x+6=12-8x\)
\(\Leftrightarrow-x+8x=12-5-6\)
\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)
a) 5-(x-6)=4(3-2x)
<=>5-x-6=12-8x
<=>-x+8x=2-5-6
<=>7x=1
<=>x=1/7
\(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)
\(\Leftrightarrow\left(\dfrac{x-3}{113}+1\right)+\left(\dfrac{x-5}{115}+1\right)=\left(\dfrac{x-7}{117}+1\right)+\left(\dfrac{x-9}{119}+1\right)\)\(\Leftrightarrow\dfrac{x+110}{113}+\dfrac{x+110}{115}=\dfrac{x+110}{117}+\dfrac{x+110}{119}\)
\(\Leftrightarrow\dfrac{x+110}{113}+\dfrac{x+110}{115}-\dfrac{x+110}{117}-\dfrac{x+110}{119}=0\)
\(\Leftrightarrow\left(x+110\right)\left(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\right)=0\)
Mà \(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\ne0\)
\(\Rightarrow x+110=0\)
\(\Rightarrow x=-110\)
\(\dfrac{x-3}{133}+\dfrac{x-5}{155}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)
\(\Leftrightarrow\left(\dfrac{x-3}{113}+1\right)+\left(\dfrac{x-5}{115}+1\right)=\left(\dfrac{x-7}{117}+1\right)+\left(\dfrac{x-9}{119}+1\right)\)
\(\Leftrightarrow\dfrac{x+130}{113}+\dfrac{x+130}{115}=\dfrac{x+130}{117}+\dfrac{x+130}{119}\)
\(\Leftrightarrow\dfrac{x+130}{113}+\dfrac{x+130}{115}-\dfrac{x+130}{117}-\dfrac{x+130}{119}=0\)
\(\Leftrightarrow\left(x+130\right)\left(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\right)=0\)
Mà \(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\ne0\)
\(\Leftrightarrow x+130=0\)
\(\Leftrightarrow x=-130\)
Vậy..
Giải:
Ta có:
\(\frac{x+1}{15}+\frac{x+2}{7}+\frac{x+4}{4}+6=0\)
\(\Leftrightarrow\frac{x}{15}+\frac{1}{15}+\frac{x}{7}+\frac{2}{7}+\frac{x}{4}+\frac{4}{4}+6=0\)
\(\Leftrightarrow\frac{x}{15}+\frac{x}{7}+\frac{x}{4}=-\frac{772}{105}\)
\(\Leftrightarrow x\left(\frac{1}{15}+\frac{1}{7}+\frac{1}{4}\right)=-\frac{772}{105}\)
\(\Leftrightarrow x=-16\)
Vậy phương trình trên có nghiệm là x = -16.
b. Cách làm tương tự.
Chúc bạn học tốt@@
\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{97}+\frac{x+100}{96}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
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
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Như vậy các bạn sẽ dễ giúp
Và cũng có nhiều bạn giúp hơn !
Bài 1.
a) ( x - 3 )( x + 7 ) = 0
<=> x - 3 = 0 hoặc x + 7 = 0
<=> x = 3 hoặc x = -7
Vậy S = { 3 ; -7 }
b) ( x - 2 )2 + ( x - 2 )( x - 3 ) = 0
<=> ( x - 2 )( x - 2 + x - 3 ) = 0
<=> ( x - 2 )( 2x - 5 ) = 0
<=> x - 2 = 0 hoặc 2x - 5 = 0
<=> x = 2 hoặc x = 5/2
Vậy S = { 2 ; 5/2 }
c) x2 - 5x + 6 = 0
<=> x2 - 2x - 3x + 6 = 0
<=> x( x - 2 ) - 3( x - 2 ) = 0
<=> ( x - 2 )( x - 3 ) = 0
<=> x - 2 = 0 hoặc x - 3 = 0
<=> x = 2 hoặc x = 3
a) Ta thấy:
\(\left(x+4\right)\left(x-4\right)=x\left(x-\frac{2}{3}\right)\)
\(\Rightarrow\left(x^2-4x\right)+\left(4x-16\right)=x^2-\frac{2}{3}x\)
\(\Rightarrow\left(x^2-16\right)-\left(4x-4x\right)=x^2-\frac{2}{3}x\)
\(\Rightarrow x^2-16-0=x^2-\frac{2}{3}x\)
\(\Rightarrow x^2-16=x^2-\frac{2}{3}x\)
\(\Rightarrow16=\frac{2}{3}x\) ( do có cùng hiệu và cùng số bị trừ )
\(\Rightarrow x=16:\frac{2}{3}\)
\(\Rightarrow x=24\)
Vậy x = 24
b.) x^3-x^2-2x=0
x(x^2-x-2)=0
x(x^2-2x+x-2)=0
x(x(x-2)+x-2)=0
x(x-2)(x+1)=0
suy ra x=0 hoặc x-2=0 hoặc x+1=0
vậy x=0 hoặc x=2 hoặc x=-1
hình như câu c đề phải là (x+4)/120 thì phải đó bạn
c.)(x+4)/120+(x+8)/116=(x+5)/119+(x+7)/117
(x+4)/120+(x+8)/116-(x+5)/119-(x+7)/117=0
(x+4)/120+1+(x+8)/116+1-(x+5)/119-1-(x+7)/117-1=0
(x+4)/120+1+(x+8)/116+1-((x+5)/119+1)-((x+7)/117+1)=0
(x+124)/120+(x+124)/116-(x+124)/119-(x+124)/117=0
(x+124)(1/120+1/116-1/119-1/117)=0
suy ra x+124=0
x=-124
3(x-1).(2x+1) - 5(x+8)(x-1)=0
(3(2x+1)-5(x+8)).(x-1)=0
(6x+3-5x-40).(x-1)=0
(x-37)(x-1)=0
=>x=37 hoac x =1
\(\frac{x+5}{105}+\frac{x+1}{109}=-\frac{x+113}{3}-\frac{x+117}{7}\)
\(\Leftrightarrow\)\(\frac{x+5}{105}+1+\frac{x+1}{109}+1=-\frac{x+113}{3}+1-\frac{x+117}{7}+1\)
\(\Leftrightarrow\)\(\frac{x+110}{105}+\frac{x+110}{109}=-\frac{x+110}{3}-\frac{x+110}{7}\)
\(\Leftrightarrow\)\(\frac{x+110}{105}+\frac{x+110}{109}+\frac{x+110}{3}+\frac{x+110}{7}=0\)
\(\Leftrightarrow\)\(\left(x+110\right)\left(\frac{1}{105}+\frac{1}{109}+\frac{1}{3}+\frac{1}{7}\right)=0\)
\(\Leftrightarrow\)\(x+110=0\) (do 1/105 + 1/109 + 1/3 + 1/7 # 0)
\(\Leftrightarrow\)\(x=-110\)
Vậy,...