tim so nguyen x biet x+5-6+8-9+3-5+9-10=9
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Tim cac so nguyen x, biet:
1. -10<x<-5
2. -4<hoac= x <hoac=3
3. n+9<x<hoac=n+12
4. n+15<x<n+15
5. x<-9 va x<9
6. |x|=|10|
7. |x|=20 va x>0
8. |x|=-100
9. |-5|+|-9|+x=14+|12|
10. 8+x=|-8|+20
tim cac so nguyen x biet
a) (2x - 10 )(x + 3)=0
b)(x+ 5)(x2 - 9)=0
\(a,\left(2x-10\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-10=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}\)
Vậy .........
\(b,\left(x+5\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x^2-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)
Vậy ......
\(a,\left(2x-10\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-10=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=10\\x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
\(b,\left(x+5\right)\left(x^2-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\x^2-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x^2=9\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=3or-3\end{cases}}}\)
bai 1 tim cac so nguyen x,y biet
a,x/3=7/y b,x/y=-3/11 c,x/y-1=5/-19
bai 2 tim cac so nguyen x,y,z,t biet
12/-6=x/5=-y/3=z/-17=-t/-9
\(a,\frac{x}{3}=\frac{7}{y}\)
\(\Rightarrow x\cdot y=3\cdot7\)
\(\Rightarrow x\cdot y=21\)
\(\Rightarrow x;y\inƯ\left(21\right)=\left\{\pm1;\pm21;\pm3;\pm7\right\}\)
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tim cac so nguyen duong x;a;b biet 4x+9=3^a;2x+5=3^b
tim so nguyen x biet a) \(3^x=9^{-6}.27^{-5}.81^8\)
b) \(2^x=8^3.8^{-10}.8^3\)
tim so nguyen x biet
a.3+x=7
b.x+5=0
c.x+9=2
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tim so nguyen x biet;
| x+9 |.2=10
|x + 9| . 2 = 10
\(\Rightarrow\) |x + 9| = 10 : 2
\(\Rightarrow\) |x + 9| = 5
\(\Rightarrow\)\(\left[\begin{matrix}x+9=5\\x+9=-5\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=5-9\\x=-5-9\end{matrix}\right.\Rightarrow}\left[\begin{matrix}x=-4\\x=-14\end{matrix}\right.\)
Vậy: x \(\in\) {-4; -14}
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\(\left|x+9\right|.2=10\)
\(\left|x+9\right|=10:2\)
\(\left|x+9\right|=5\)
\(\Rightarrow\left\{\begin{matrix}x+9=5\\-\left(x+9\right)=5\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=-4\\-x-9=5\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=-4\\-x=14\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=-4\\x=-14\end{matrix}\right.\)
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l x+9 l.2=10
\(\Rightarrow\)lx+9l=10:2
\(\Rightarrow\)lx+9l=5 \(\Rightarrow\)\(\left\{\begin{matrix}x+9=5\\x+9=-5\end{matrix}\right.\) \(\Rightarrow\)\(\left\{\begin{matrix}x=-4\\x=-14\end{matrix}\right.\) vay x\(\in\){-4;-14}
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tim so nguyen x
(3x+9).(3x-6)=0(2x+15)-25=47-(10-x)30(x=2)-6(x-5)-24x=100/4-3x/=8/2x-5/=13/7x+3/=66
1) (3x + 9)(3x - 6) = 0
=> \(\orbr{\begin{cases}3x+9=0\\3x-6=0\end{cases}}\)
=> \(\orbr{\begin{cases}3x=-9\\3x=6\end{cases}}\)
=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Vậy ...
b) (2x + 15) - 25 = 47 - (10 - x)
=> 2x - 10 = 37 + x
=> 2x - x = 37 + 10
=> x = 47
3, tương tự
4) |4 - 3x| = 8
=> \(\orbr{\begin{cases}4-3x=8\\4-3x=-8\end{cases}}\)
=> \(\orbr{\begin{cases}3x=-4\\3x=12\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{4}{3}\\x=4\end{cases}}\)
Vì x là số nguyên nên ...
còn lại tương tự
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tim so nguyen x biet
a,9-25=[7-x]-[25+7]
b,-26-[x-7]=0
c,30+[32-x]=10
d,2.x-18=10
e,3.x+26=5
f,/x/-5=-12+30
g,[/x/+1].[4-2x]=0
h,8+/x/=/-8/+11
i,4.[x+1]-[3x+1]=14
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