TIM X BIET
(2X-3/5)^2-9/25=0
Tim xthuoc Z biet:
1,|2x-5|-|2x+9|=0
2,|x+1|-|x+2|-|3-x|=7
3,|2x+3|+|3x+2|-|4-x|=10
tim x thuoc z
1)26-|x+9|=-13
2)|x+7|-13=25
tim x biet
1)123-3.(x+4)=23
2)720:[41-(2x-5)]=23.5
Tìm x thuoc z:
1) \(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)\)
\(\Leftrightarrow\left|x+9\right|=39\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=39-9=30\\x=-39-9=-48\end{matrix}\right.\)
Vậy: \(x\in\left\{30;-48\right\}\)
2) \(\left|x+7\right|-13=25\)
\(\Leftrightarrow\left|x+7\right|=25+13=38\)
\(\Leftrightarrow x+7\in\left\{38;-38\right\}\)
\(\Leftrightarrow x\in\left\{31;-45\right\}\)
Vậy:.................
tim x biet
\(1)123-3.\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=123-23\)
\(\Leftrightarrow3\left(x+4\right)=100\)
\(\Leftrightarrow x+4=\frac{100}{3}\)
\(\Leftrightarrow x=\frac{100}{3}-4=\frac{100-12}{3}=\frac{88}{3}\)
Vậy:................
2) Tương tự
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 ......
\(a,\left(2x-10\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-10=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=10\\x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
\(b,\left(x+5\right)\left(x^2-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\x^2-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x^2=9\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=3or-3\end{cases}}}\)
Tim x biet
(x+1/5)-4=-2
(2x+3)*(x-7)=0
31/9(x)-5/2=8/3
Ta có : \(\left(2x+3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=-3\\x=7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=7\end{cases}}\)
Tìm x,biết :
\(a,\left(x+\frac{1}{5}\right)-4=-2\)
\(\left(x+\frac{1}{5}\right)=2\)
\(x+\frac{1}{5}=2\)
\(x=\frac{9}{5}\)
b,\(\left(2x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\x-7=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=7\end{cases}}\)
\(c,\frac{31}{9}x-\frac{5}{2}=\frac{8}{3}\)
\(\frac{31}{9}x=\frac{8}{3}+\frac{5}{2}\)
\(\frac{31}{9}x=\frac{31}{6}\)
\(x=\frac{3}{2}\)
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
bai 1)tim x biet
a )24-(36+5)=x
b)14-21=(13-x)-(15+8)
bai 2) tim x biet
a)17-x=-25+(-16+9)
b)3x-21=-19-(-2x)
Bài 1: Tìm x, biết
a )24-(36+5)=x b)14-21=(13-x)-(15+8)
24-41=x (13-x)-23=-7
x=-17 13-x=(-7)+23
Vậy x=-17 13-x=16
x=13-16
x=-3 Vậy x=-3
Bài 2:Tìm x, biết
a)17-x=-25+(-16+9) b)3x-21=-19-(-2x)
17-x=-25+(-7) 3x-21=-19+2x
17-x=-32 3x-2x=-19+21
x=17-(-32) x=4
x=49 Vậy x=4
Vậy x=49
Bài 1:
a. 24 - (36+5) = x
=> 24 - 41 = x
=> -17 = x
=> x = -17
b. 14 - 21 = (13 - x) - (15 + 8)
=> -7 = 13 - x - 23
=> -7 - 13 + 23 = -x
=> 3 = -x
=> x = -3
Bài 2:
a. 17 - x = -25 + (-16 + 9)
=> 17 - x = -25 + (-7)
=> 17 - x = -32
=> 17 + 32 = x
=> x = 49
b. 3x - 21 = -19 - (-2x)
=> 3x - 21 = -19 + 2x
=> 3x - 2x = -19 + 21
=> x = 2
Bài1
a)24-(36+5)=x\(\Rightarrow\)x=24-41=-17
b)14-21=(13-x)-(15+8)\(\Rightarrow\)(13-x)-(15+8)=-7\(\Rightarrow\)13-x=-7+23=16\(\Rightarrow\)x=13-16=-3
Bài2
a)17-x=-25+(-16+9)\(\Rightarrow\)17-x=-25+-7\(\Rightarrow\)17-x=-32\(\Rightarrow\)x=17-(-32)=17+32=49
b)3x-21=-19-(-2x)\(\Rightarrow\)3x-(-2x)=-19-21\(\Rightarrow\)3x+2x=-40\(\Rightarrow\)5x=-40\(\Rightarrow\)x=-8
Tim cac so nguyen x biet:
a,(x^2--5)(x^2-25)<0
b,(x-2)(x^2+1)=0
c,(x+3)(x^2+9)<0
d,(x-1)92x^2-8)^2=3
tim x biet
a. (x-3)(x+5)=0
b./9-2x/+x+3=2x+15
c./x+1/+/x-1/=4
a/ (x - 3)(x + 5) = 0
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
b/ |9 - 2x| + x + 3 = 2x + 15
=> |9 - 2x| = 2x + 15 - 3 - x
=> |9 - 2x| = x + 12
\(\Rightarrow\orbr{\begin{cases}9-2x=x+12\\9-2x=-x-12\end{cases}}\Rightarrow\orbr{\begin{cases}-2x-x=12-9\\-2x+x=-12-9\end{cases}}\Rightarrow\orbr{\begin{cases}-3x=3\\-x=-21\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=21\end{cases}}\)
c/ TH1: Nếu x > 1 thì x + 1 + x - 1 = 4
=> 2x = 4
=> x = 2
TH2: Nếu x < 1 thì x + 1 + x - 1 = -4
=> 2x = -4
=> x = -2
TH3: Nếu x = 1 thì 1 + 1 + 1 - 1 = 4 (vô lí)
Vậy x = 2 hoặc x = -2
Câu c thì mình không chắc cho lắm, không biết có đúng không nữa. ._.
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
tim x
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(\Rightarrow x=0\)