tìm x
4.x-x+18=600
5.x+2.x+140=700
K
Khách
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D
17 tháng 1 2023
`|2x-3|+|x-1|=6005`
`@TH1: x < 1`
`=>3-2x+1-x=6005`
\(<=>x=\dfrac{-6001}{3}\) (t/m)
\(@TH2: 1 \le x < \dfrac{3}{2}\)
`=>3-2x+x-1=6005`
`<=>x=-6003` (ko t/m)
`@TH3: x \ge \dfrac{3}{2}\)
`=>2x-3+x-1=6005`
`<=>x=2003` (t/m)
PH
0
C
1
L
14 tháng 8 2021
a) \(x^2-x+x=4\)
\(x^2=4\)
\(x=\pm2\)
b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)
\(\left(x-5\right)\left(3x-2\right)=0\)
\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c) Ta có: \(a+b+c=5-3-2=0\)
\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)
d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :
\(t^2-11t+18=0\)
\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)
\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)
\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)
C
2
14 tháng 8 2021
a:Ta có: \(x\left(x-1\right)+x=4\)
\(\Leftrightarrow x^2-x+x=4\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
b: Ta có: \(3x\left(x-5\right)-2x+10=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c: Ta có: \(5x^2-3x-2=0\)
\(\Leftrightarrow5x^2-5x+2x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d: Ta có: \(x^4-11x^2+18=0\)
\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)
\(\Leftrightarrow x^2\left(x^2-9\right)-2\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
14 tháng 8 2021
a) x(x-1)+x=4
⇔x2=4⇔\(x=\pm2\)
b)3x(x-5)-2x+10=0
⇔3x(x-5)-2(x-5)=0
⇔(x-5)(3x-1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)
c)5x2-3x-2=0
⇔ 5x(x-1)+2(x-1)=0
⇔ (x-1)(5x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d)x4-11x2+18=0
⇔ x2(x2-2)-9(x2-2)=0
⇔ (x2-2)(x2-9)=0
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)
PT
1
CM
17 tháng 8 2019
a) x − 3 49 = − 2 7 ⇒ x − 3 49 = − 14 49 ⇒ x = − 11
b) x 4 = x + 1 8 ⇒ 2 x 4 = x + 1 8 ⇒ x = 1
H
26 tháng 12 2022
Giải
Bài 1
\(1,a,-40-37-\left(-29\right)\\ =-40-37+29\\ =-77+29=-48\\ b,27+\left(-36\right)-\left(-45\right)=27-36+45\\ =-9+45\\ =36\\ c,-125-\left[\left(-18\right)-125\right]\\ =-125+18+125\\ =\left(-125+125\right)+18\\ =18\\ d,140+\left[\left(-184\right)-140\right]\\ =140-184-140\\ =\left(140-140\right)-184\\ =0-184\\ =-184\)
Bài 2
\(a,x-\left(-15\right)=-8\\ =>x+15=-8\\ =>x=-8-15\\ =>x=-23\\ b,-40-x=-35\\ =>x=\left(-40\right)-\left(-35\right)\\ =>x=-40+35\\ =>x=-5\\ c,x+\left(-50\right)=-27\\ =>x-50=-27\\ =>x=-27+50\\ =>x=23\)
NK
26 tháng 12 2022
1, Thực hiện phép tính:
\(a,-40+\left(-37\right)-\left(-29\right)=-77-\left(-29\right)=-48\)
\(b,27+\left(-36\right)-\left(-45\right)=-9-\left(-45\right)=36\)
\(c,-125-\left[\left(-18\right)-125\right]=-125-\left(-143\right)=18\)
\(d,140+\left[\left(-184\right)-140\right]=140+\left(-324\right)=-184\)
2, Tìm x biết:
\(a,x-\left(-15\right)=-8\)
\(x=-8+\left(-15\right)\)
\(x=-23\)
\(b,-40-x=-35\)
\(x=-40-\left(-35\right)\)
\(x=-5\)
\(c,x+\left(-50\right)=-27\)
\(x=-27-\left(-50\right)\)
\(x=23\)
PT
1
H9
HT.Phong (9A5)
CTVHS
7 tháng 10 2023
a) Do x chia hết cho 40 và chia hết cho 50 nên:
\(x\in BC\left(40,50\right)\)
Ta có:
\(B\left(40\right)=\left\{0;40;80;120;160;200;240;280;320;360;400;440;480;520;..\right\}\)
\(B\left(50\right)=\left\{0;50;100;150;200;250;300;350;400;450;500;550...\right\}\)
\(\Rightarrow BC\left(40,50\right)=\left\{0;200;400;600;...\right\}\)
Mà: \(x< 500\)
\(\Rightarrow x\in\left\{0;200;400\right\}\)
b) A chia hết cho 140 và A chia hết cho 350 nên:
\(\Rightarrow A\in BC\left(140,350\right)\)
Ta có:
\(B\left(140\right)=\left\{0;140;280;420;560;700;840;980;1120;1260;1400;1540\right\}\)
\(B\left(350\right)=\left\{0;350;700;1050;1400;1750;...\right\}\)
\(\Rightarrow BC\left(140;350\right)=\left\{0;700;1400;...\right\}\)
Mà: \(1200< A< 1500\)
\(\Rightarrow A\in\left\{1400\right\}\)
bạn muốn tôi chả lời cái gì
\(4x-x+18=600\Leftrightarrow3x+18=600\Leftrightarrow3x=600-18\)
\(\Leftrightarrow3x=582\Leftrightarrow x=582:3\Leftrightarrow x=194\)
\(5x+2x+140=700\Rightarrow5x+2x=700-140\)
\(5x+2x=560\Leftrightarrow x\left(5+2\right)=560\)
\(7x=560\Leftrightarrow x=560:7\)
\(x=80\)