tìm x biết
a)565-(6x+70)=435
b)100-7(x-5)=58
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
24 tháng 9 2021
\(a,\Rightarrow6x+70=130\Rightarrow6x=60\Rightarrow x=10\\ b,\Rightarrow240=\left(x+70\right):14-40\\ \Rightarrow\left(x+70\right):14=280\\ \Rightarrow x+70=3920\Rightarrow x=3850\\ c,\Rightarrow x-15=75\Rightarrow x=90\\ d,\Rightarrow\left(x+175\right):5=680-30=650\\ \Rightarrow x+175=3250\Rightarrow x=3075\\ e,\Rightarrow x-4867=1004523\Rightarrow x=1009390\\ f,\Rightarrow x+32-17=24\Rightarrow x=9\\ g,\Rightarrow3\left(x+1\right)=54\Rightarrow x+1=18\Rightarrow x=17\\ h,\Rightarrow19\left(35:x+3\right)=152\\ \Rightarrow35:x+3=8\Rightarrow35:x=11\Rightarrow x=\dfrac{35}{11}\)
L
4
I
21 tháng 2 2020
a ) 100-7(x-5)=58
=> 7(x-5)=42
=.> x-5 = 6
=> x=11
180-5(x+1)=30
=> 5(x+1)=150
=> x+1=30\
=> x=29
6x-5=613
=> 6x=618
=> x=103
L
21 tháng 2 2020
a) \(100-7\left(x-5\right)=58\)
\(\Leftrightarrow7\left(x-5\right)=42\)
\(\Leftrightarrow x-5=6\)
\(\Leftrightarrow x=11\)
b) \(180-5\left(x+1\right)=30\)
\(\Leftrightarrow5\left(x+1\right)=150\)
\(\Leftrightarrow x+1=30\)
\(\Leftrightarrow x=29\)
c) \(6x-5=613\)
\(\Leftrightarrow6x=618\)
\(\Leftrightarrow x=103\)
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 \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
6 tháng 7 2021
\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)
\(\Leftrightarrow-5x-18=0\)
\(\Leftrightarrow x=-\dfrac{18}{5}\)
Vậy ...
\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)
\(\Leftrightarrow12x+6=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy ...
\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)
\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)
Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)
\(\Rightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy ...
\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)
Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy ...
\(a,565-\left(6x+70\right)=435\)
\(6x+70=565-435\)
\(6x+70=130\)
\(6x=130-70\)
\(6x=60\)
\(x=10\)
\(b,100-7\left(x-5\right)=58\)
\(100-7x+35=58\)
\(100-7x=58-35\)
\(100-7x=23\)
\(7x=100-23\)
\(7x=77\)
\(x=11\)
a, 565-(6x+70)=435
6x+70=565-435
6x+70=130
6x=130-70
6x=60
x=60:6
x=6
b, 100-7(x-5)=58
7(x-5)=100-58
7(x-5)=42
x-5=42:7
x-5=6
x=6+5
x=11