Tìm X e N biết
a) 29 + 2x = 59
b) 2020-6(17-x)2002
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12 tháng 7 2021
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
NK
0
14 tháng 4 2020
Bài 1: Tính
a) Ta có: \(\left(-25\right)\cdot68+\left(-34\right)\cdot\left(-250\right)\)
\(=-25\cdot68+\left(-340\right)\cdot\left(-25\right)\)
\(=-25\cdot\left(68-340\right)\)
\(=-25\cdot\left(-272\right)\)
\(=6800\)
b) Ta có: \(1999+\left(-2000\right)+2001+\left(-2002\right)\)
\(=1999-2000+2001-2002\)
\(=-1-1=-2\)
c) Ta có: \(515+\left[72+\left(-515\right)+\left(-32\right)\right]\)
\(=515+72-515-32\)
\(=40\)
d) Ta có: \(\left(2736-75\right)-2736+175\)
\(=2736-75-2736+175\)
\(=100\)
e) Ta có: \(-2020-\left(157-2020\right)-\left(-257\right)\)
\(=-2020-157+2020+257\)
\(=100\)
Bài 2: Tìm x
a) Ta có: \(x-\left|-2\right|=\left|-18\right|\)
\(\Leftrightarrow x-2=18\)
hay x=20
Vậy: x=20
b) Ta có: \(2x-\left|+14\right|=\left|-14\right|\)
\(\Leftrightarrow2x-14=14\)
\(\Leftrightarrow2x=28\)
hay x=14
Vậy: x=14
c) Ta có: \(\left|x+4\right|+5=20-\left(-12-7\right)\)
\(\Leftrightarrow\left|x+4\right|+5=20+12+7\)
\(\Leftrightarrow\left|x+4\right|=39-5=34\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=34\\x+4=-34\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-38\end{matrix}\right.\)
Vậy: x∈{30;-38}
d) Ta có: \(15-\left|2-x\right|=\left(-2\right)^2\)
\(\Leftrightarrow15-\left|2-x\right|=4\)
\(\Leftrightarrow\left|2-x\right|=11\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=11\\2-x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=13\end{matrix}\right.\)
Vậy: x∈{-9;13}
e) Ta có: \(\left|15-x\right|+\left|-25\right|=\left|-55\right|\)
\(\Leftrightarrow\left|15-x\right|+25=55\)
\(\Leftrightarrow\left|15-x\right|=30\)
\(\Leftrightarrow\left[{}\begin{matrix}15-x=30\\15-x=-30\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-15\\x=45\end{matrix}\right.\)
Vậy: x∈{-15;45}
g) Ta có: \(\left|17-\left(-4\right)\right|+\left|-24-\left(-5\right)\right|=\left|-x+3\right|\)
\(\Leftrightarrow\left|17+4\right|+\left|-24+5\right|=\left|3-x\right|\)
\(\Leftrightarrow\left|3-x\right|=40\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=40\\3-x=-40\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-37\\x=43\end{matrix}\right.\)
Vậy: x∈{-37;43}
LQ
3
PN
4
21 tháng 1 2022
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
a) 29 + 2x = 59
2x = 59 - 29
2x = 30
x = 30 : 2
x = 15
b) 2020 - 6(17-x) = 2002
6(17-x) = 2020 - 2002
6(17-x) = 18
17-x = 18 : 6
17-x = 3
x = 17 - 3
x = 14