Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
dẽ qua ak nhưng giúp mình làm bài này đi
cho tam giac abc . co canh bc=12cm, duong cao ah=8cm
a> tinh s tam giac abc
b> tren canh bc lay diem e sao cho be=3/4bc. tinh s tam giac abe va s tam giac ace ( bằng nhiều cách
c> lay diem chinh giua cua canh ac va m . tinh s tam giac ame
\(\frac{x+1}{58}+\frac{x+2}{57}=\frac{x+3}{56}+\frac{x+4}{55}\)
\(\Rightarrow\left(\frac{x+1}{58}+1\right)+\left(\frac{x+2}{57}+1\right)=\left(\frac{x+3}{56}+1\right)+\left(\frac{x+4}{55}+1\right)\)
\(\Rightarrow\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)
\(\Rightarrow\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)
\(\Rightarrow\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)
Mà \(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\ne0\)
\(\Rightarrow x+59=0\)
\(\Rightarrow x=-59\)
a) Ta có: \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow-2x+3+x+4=0\)
\(\Leftrightarrow-x+7=0\)
\(\Leftrightarrow-x=-7\)
hay x=7
Vậy: S={7}
b) Ta có: \(\dfrac{2+x}{5}-0.5x=\dfrac{1-2x}{4}+0.25\)
\(\Leftrightarrow\dfrac{4\left(2+x\right)}{20}-\dfrac{0.5x\cdot20}{20}=\dfrac{5\left(1-2x\right)}{20}+\dfrac{20\cdot0.25}{20}\)
\(\Leftrightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)
\(\Leftrightarrow8+4x-10x=5-10x+5\)
\(\Leftrightarrow-6x+8=-10x+10\)
\(\Leftrightarrow-6x+8+10x-10=0\)
\(\Leftrightarrow4x-2=0\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
d) Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-59}{1}+\dfrac{x-58}{2}+\dfrac{x-57}{3}\)
\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}-1=\dfrac{x-59}{1}-1+\dfrac{x-58}{2}-1+\dfrac{x-57}{3}-1\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{1}+\dfrac{x-60}{2}+\dfrac{x-60}{3}\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}\right)-\left(x-60\right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
mà \(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\ne0\)
nên x-60=0
hay x=60
Vậy: S={60}
(x -1)/59 -1 +(x-2)/58 -1 +(x-3)/57 -1 = (x-3)/56 -1 +(x-4)/55 -1 +(x-5)/54 -1
<=> (x-60)/59 +(x-60)/58 + (X-60)/57 -(x-60)/56 - (X-60)/55 -(X-60)/54 =0
<=> (x-60).(1/59 +1/58 +1/57 -1/56 -1/55 - 1/54)=0
vì 1/59 +1/58 +1/57 -1/56 -1/55 -1/54 <0
nên x-60 =0 <=> x=60
đề bài của bạn bi sai vì vế trái không thể bằng vế phải nếu đề đúng thì phải là :
(x-1)/59 +(x-2)/58 +(x-3)/57 =(x-4)/56 +(x-5)/55 +(x-6)/54
khí đó bạn giải cách như trên ,chúc bạn học toán tốt
\(\frac{x+1}{58}+\frac{x+2}{57}=\frac{x+3}{56}+\frac{x+4}{55}\)
\(\Rightarrow\frac{x+1}{58}+1+\frac{x+2}{57}+1=\frac{x+3}{56}+1+\frac{x+4}{55}+1\)
\(\Rightarrow\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)
\(\Rightarrow\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)
\(\Rightarrow\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)
Mà \(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\ne0\)
\(\Rightarrow x+59=0\)
\(\Rightarrow x=-59\)
Mình khuyên bạn thế này :
Bạn nên tách những câu hỏi ra
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
\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)
\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)
\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)
\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)
\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
3.15:
a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)
b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3.16
\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)
\(\Leftrightarrow-14m+35-2m^2+8=0\)
\(\Leftrightarrow-14m-2m^2+43=0\)
\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)
\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)
\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)
\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)
pt vô nghiệm
Bài 1:
\(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
\(\Rightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)
\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)
\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)
\(\Rightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)
Mà \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)
\(\Rightarrow x+95=0\)
\(\Rightarrow x=-95\)
Vậy x = -95
Bài 2: tương tự
\(\frac{x+1}{94}\)+\(\frac{x+2}{93}\)+\(\frac{x +3}{92}\)= \(\frac{x+4}{91}\)+ \(\frac{x+5}{90}\)+ \(\frac{x+6}{89}\)
<=> [(\(\frac{x+1}{94}\)+\(\frac{x+2}{93}\)+\(\frac{x+3}{92}\)+3)]= [(\(\frac{x+4}{91}\)+\(\frac{x+5}{90}\)+\(\frac{x+6}{89}\)+3)]
<=> [(\(\frac{x+1}{94}\)+1)+(\(\frac{x+2}{93}\)+1)+(\(\frac{x+3}{92}\)+1)]- [(\(\frac{x+4}{91}\)+1)+(\(\frac{x+5}{90}\)+1)+(\(\frac{x+6}{89}\)+1)] =0
<=> [(\(\frac{x+1}{94}\)+ \(\frac{94}{94}\))+(\(\frac{x+2}{93}\)+\(\frac{93}{93}\))+(\(\frac{x+3}{92}\)+\(\frac{92}{92}\))] -[(\(\frac{x+4}{91}\)+\(\frac{91}{91}\))+(\(\frac{x+5}{90}\)+\(\frac{90}{90}\))+(\(\frac{x+6}{89}\)+\(\frac{89}{89}\))] =0
<=> (\(\frac{x+95}{94}\)+\(\frac{x+95}{93}\)+\(\frac{x+95}{92}\)) -(\(\frac{x+95}{91}\)+\(\frac{x+95}{90}\)+\(\frac{x+95}{89}\)) =0
<=> (x+95)( \(\frac{1}{94}\)+\(\frac{1}{93}\)+\(\frac{1}{92}\)-\(\frac{1}{91}\)-\(\frac{1}{90}\)-\(\frac{1}{89}\)) =0
Vì (\(\frac{1}{94}\)+\(\frac{1}{93}\)+\(\frac{1}{92}\)-\(\frac{1}{91}\)-\(\frac{1}{90}\)-\(\frac{1}{89}\)) \(\ne\) 0
=> x+95=0
<=> x= -95
Vậy S={-95}
Ta có:
⇔ x - 64 = 0 ⇔ x = 64
Vậy phương trình đã cho có nghiệm là x = 64