184654 x 109
Câu 3. Có thể chuyển biểu thức 109 x 7 + 109 x 3 thành biểu thức: *
A. 7 x 109 + 3
B. 109 x 3 + 7
C. 109 x 7 + 3
D. 109 x (7 + 3)
giá trị x<0 thỏa mãn
x/109= -109/-x
Từ đề bài , ta có :
(-x).x = (-109).109
-x2 = -1092
=> x2 = 109
=> x = {109 ; -109}
MÀ x < 0
=> x = -109
109×(200-x)=1092
109x(200-x)=11881
( 200-x) = 11881:109
200-x =109
x = 200 - 109
x = 91
chúc bạn học tốt
tìm x :\(\frac{x+14}{200}+\frac{x+27}{187}+\frac{x+105}{109}=\frac{x+200}{14}+\frac{x+187}{27}+\frac{x+109}{105}\)
\(\left(\frac{x+14}{200}+1\right)+\left(\frac{x+27}{187}+1\right)+\left(\frac{x+105}{109}+1\right)=\left(\frac{x+200}{14}+1\right)\)
\(+\left(\frac{x+187}{27}+1\right)+\left(\frac{x+109}{105}+1\right)\)
\(\Rightarrow\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\)
\(\Rightarrow\left(x+214\right)\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)=0\)
Mà \(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\ne0\)
\(\Rightarrow x+214=0\)
\(\Rightarrow x=-214\)
Vậy x = -214
a)5-(13-3x)=109-(32+109)
b)/3x-6/+(x-2^2)=0
c)/7-2x/=-13-5.(-8)
a)5-(13-3x)=109-(32+109)
5-(13-3x)=-32
13-3x =37
3x =-24
3x =-6
a)5-(13-3x)=109-(32+109)
5-13+3x =109-32-109
-8+3x=-32
3x=-32-(-8)
3x=-24
x=-24:3
x=-8
b)|3x-6|+(x-22)=0
(3x-6)+(x-22)=0
3x-6+x-22=0
3x+x=0+6+22
4x=10
x=\(\dfrac{5}{2}\)
c)/7-2x/=-13-5.(-8)
|7-2x|=27
⇒7-2x=27 hoặc 7-2x=-27
x=-10 hoặc x=-17
Tìm x bt
a) (x-5) + (x-7) = -12
b) 3x - 44 = -109 -2x
c) 2 ( x-5) - 5 (3+x) =-109
d) (x-3) . (x+5 ) < 0
\(a.\)
\(\Leftrightarrow x-5+x-7=-12.\)
\(\Leftrightarrow x=-12+-12\)
\(\Leftrightarrow x=-24\)
\(b.\)
\(\Leftrightarrow3x+2x=-109+44\)
\(\Leftrightarrow5x=-65\)
\(\Rightarrow x=-13\)
\(c.\)
\(\Leftrightarrow2x-10-15+5x=-109\)
\(\Leftrightarrow2x-5x=-109+15+10\)
\(\Leftrightarrow3x=-84\)
\(\Rightarrow x=-28\)
\(d.\)
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d) \(\left(x-3\right)\left(x+5\right)< 0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3< 0\\x+5< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x< 3\\x< -5\end{cases}}\)
Giúp mình nha
\(\frac{x+14}{200}+\frac{x+27}{187}+\frac{x+105}{109}=\frac{x+200}{14}+\frac{x+187}{27}+\frac{x+109}{105}\)
\(\Leftrightarrow\frac{x+14}{200}+\frac{x+27}{187}+\frac{x+105}{109}-\frac{x+200}{14}-\frac{x+187}{27}-\frac{x+109}{105}=0\)
\(\Leftrightarrow\left(\frac{x+14}{200}+1\right)+\left(\frac{x+27}{187}+1\right)+\left(\frac{x+105}{109}+1\right)-\left(\frac{x+200}{14}+1\right)-\left(\frac{x+187}{27}+1\right)-\left(\frac{x+109}{105}+1\right)=0\)\(\Leftrightarrow\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\)
\(\Leftrightarrow\left(x+214\right)\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)=0\)
Mà \(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}< \frac{1}{14}+\frac{1}{27}+\frac{1}{105}\Rightarrow\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\ne0\)
\(\Rightarrow x+214=0\)
\(\Rightarrow x=-214\)
Vậy x=-214
\(\frac{x+14}{200}+\frac{x+27}{187}+\frac{x+105}{109}=\frac{x+200}{14}+\frac{x+187}{27}+\frac{x+109}{105}\\\Leftrightarrow \frac{x+14}{200}+1+\frac{x+27}{187}+1+\frac{x+105}{109}+1=\frac{x+200}{14}+1+\frac{x+187}{27}+1+\frac{x+109}{105}+1\\\Leftrightarrow \frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\\\Leftrightarrow \left(x+214\right)\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)=0\)
\(\Leftrightarrow x+214=0\left(vi\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\ne0\right)\\\Leftrightarrow x=-214 \)
Vậy tập nghiệp của phương trình trên là \(S=\left\{-214\right\}\)
Tìm :
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\(x\) + \(\dfrac{9}{10}\) = \(\dfrac{16}{10}\)
\(x\) = \(\dfrac{16}{10}\) - \(\dfrac{9}{10}\)
\(x\) = \(\dfrac{7}{10}\)
\(x\) =0,7
109*x+137*x=75030
109 . x + 137 . x = 75030
x . ( 109 + 137 ) = 75030
x . 246 = 75030
x = 75030 : 246
x = 305