Tìm x biết:
a, x : 25 = 1205 : 5
b, 2018 – x = 425 +725
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) PT \(\Leftrightarrow x^2-x-x^2+2x=5\) \(\Rightarrow x=5\)
Vậy ...
b) PT \(\Leftrightarrow8x=16\) \(\Rightarrow x=2\)
Vậy ...
a: Ta có: \(x\left(x-1\right)-x^2+2x=5\)
\(\Leftrightarrow x^2-x-x^2+2x=5\)
hay x=5
b: Ta có: \(2x\left(3x+4\right)-6x^2=16\)
\(\Leftrightarrow6x^2+8x-6x^2=16\)
\(\Leftrightarrow8x=16\)
hay x=2
a, 35 - \(x\) < 35 - 5
35 - \(x\) < 30
\(x\) < 35 - 30
\(x\) < 5
\(x\) = 1; 2; 3; 4; 5
b, \(x\) - 10 < 35 - 10
\(x\) - 10 < 15
\(x\) < 15 + 10
\(x\) < 25
\(x\) = 0; 1; 2; 3; 4; 5; ....;24
c, \(x\) - 10 < 45
\(x\) < 45 + 10
\(x\) < 55
\(x\) = 0; 1; 2; 3; 4;...;54
a) x3-1-(x2+2x)(x-2)=5
⇔ x3-1-x3+4x=5
⇔ 4x=6
⇔ \(x=\dfrac{3}{2}\)
a: =>x/-3=3
hay x=-9
b: =>x/9=-1/9
hay x=-1
c: =>x+1/5=-1/3
hay x=-8/15
d: =>-7/x=-7/9
hay x=9
a, \(\dfrac{x}{-3}=3\Leftrightarrow x=-9\)
b, \(\dfrac{x}{9}=-\dfrac{1}{9}\Rightarrow x=-1\)
c, \(\dfrac{x+3}{15}=-\dfrac{6}{15}\Rightarrow x=-9\)
d, \(\dfrac{42}{-54}=-\dfrac{42}{6x}\Rightarrow6x=54\Leftrightarrow x=9\)
a) \(3\left(x-2\right)+2\left(x-3\right)=5\)
\(\Rightarrow3x-6+2x-6=5\)
\(\Rightarrow5x=17\Rightarrow x=\dfrac{17}{5}\)
b) \(\left(2x-8\right)^2-16=0\)
\(\Rightarrow\left(2x-8-4\right)\left(2x-8+4\right)=0\)
\(\Rightarrow\left(2x-12\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=12\\2x=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)
c) \(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=3\)
\(\Rightarrow4x^2-4x+1-4x^2+12x-x+3=3\)
\(\Rightarrow7x=-1\Rightarrow x=-\dfrac{1}{7}\)
a: Ta có: \(3\left(x-2\right)+2\left(x-3\right)=5\)
\(\Leftrightarrow3x-6+2x-6=5\)
\(\Leftrightarrow5x=17\)
hay \(x=\dfrac{17}{5}\)
b: Ta có: \(\left(2x-8\right)^2-16=0\)
\(\Leftrightarrow\left(2x-4\right)\left(2x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
a) Ta có: \(x\left(x-1\right)-x^2+2x=5\)
\(\Leftrightarrow x^2-x-x^2+2x=5\)
hay x=5
b) Ta có: \(2x^2-2x=\left(x-1\right)^2\)
\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) Ta có: \(\left(x+3\right)\cdot\left(x^2-3x+9\right)-x\left(x-2\right)^2=19\)
\(\Leftrightarrow x^3+27-x\left(x^2-4x+4\right)-19=0\)
\(\Leftrightarrow x^3+8-x^3+4x^2-4x=0\)
\(\Leftrightarrow4x^2-4x+8=0\)(Vô lý)
a: \(\sqrt{\left(x+1\right)^2}=5\)(ĐKXĐ: \(x\in R\))
=>|x+1|=5
=>\(\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)
b: Sửa đề: \(5\sqrt{9x-9}-\sqrt{4\left(x-1\right)}+\sqrt{36\left(x-1\right)}-18=0\)
ĐKXĐ: x>=1
\(PT\Leftrightarrow5\cdot3\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}-18=0\)
=>\(15\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}=18\)
=>\(19\sqrt{x-1}=18\)
=>\(\sqrt{x-1}=\dfrac{18}{19}\)
=>\(x-1=\left(\dfrac{18}{19}\right)^2=\dfrac{324}{361}\)
=>\(x=\dfrac{324}{361}+1=\dfrac{324+361}{361}=\dfrac{685}{361}\)
Lời giải:
a. PT $\Leftrightarrow |x+1|=5$
$\Leftrightarrow x+1=\pm 5\Leftrightarrow x=4$ hoặc $x=-6$
b. ** Sửa $x-9$ thành $x-1$
ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow 5\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}-18=0$
$\Leftrightarrow (5-2+6)\sqrt{x-1}=18$
$\Leftrightarrow 9\sqrt{x-1}=18$
$\Leftrightarrow \sqrt{x-1}=2$
$\Leftrightarrow x-1=4$
$\Leftrightarrow x=5$ (tm)
a: (x-2)(y-3)=5
=>\(\left(x-2\right)\cdot\left(y-3\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
=>\(\left(x-2;y-3\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(3;8\right);\left(7;4\right);\left(1;-2\right);\left(-3;2\right)\right\}\)
b: (2x-1)*(y-4)=-11
=>\(\left(2x-1\right)\cdot\left(y-4\right)=1\cdot\left(-11\right)=\left(-11\right)\cdot1=\left(-1\right)\cdot11=11\cdot\left(-1\right)\)
=>\(\left(2x-1;y-4\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(-5;5\right);\left(0;15\right);\left(6;3\right)\right\}\)
c: xy-2x+y=3
=>\(x\left(y-2\right)+y-2=1\)
=>\(\left(x+1\right)\left(y-2\right)=1\)
=>\(\left(x+1\right)\cdot\left(y-2\right)=1\cdot1=\left(-1\right)\cdot\left(-1\right)\)
=>\(\left(x+1;y-2\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;3\right);\left(-2;1\right)\right\}\)
a: =1250:25=50
b: =234(45+55)+234=23400+234=23634
c: =1000x1000=1000000
a, x : 25 = 1205 : 5
x : 25 = 241
x = 241 x 25
x = 6025
b, 2018 – x = 425 +725
2018 – x = 1150
x = 2018 – 1150
x = 868