Tìm x biết: 4+12+20+36+...+x=256
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(x+4)+(x+8)+(x+12)+...+(x+32)=256
=>(x+x+x+......+x)+(4+8+12+.......+32)=256
mà số số hạng x tương ứng với số số hạng 4+8+12+...+32
=> có (32-4):4+1=8 số x
và tổng 4+8+12+.....+32=(32+4)x8:2=144
thay vào ta có :
8x X+144=256
=>8xX=112
=>X=14
1: \(\Leftrightarrow x=UCLN\left(24;36;150\right)=6\)
2: \(\Leftrightarrow x\in\left\{24;48;72;...\right\}\)
mà 16<=x<=50
nên \(x\in\left\{24;48\right\}\)
3: \(\Leftrightarrow x\inƯ\left(6\right)\)
mà x>-10
nên \(x\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
4: \(\Leftrightarrow x\in BC\left(4;5;8\right)\)
\(\Leftrightarrow x\in\left\{...;-40;0;40;80;120;160;200;...\right\}\)
mà -20<x<180
nên \(x\in\left\{0;40;80;120;160\right\}\)
A) \(2x+12=36\)
\(\Leftrightarrow2x=36-12\)
\(\Leftrightarrow2x=24\)
\(\Leftrightarrow x=12\)
B) \(\left(x+21\right):8+12=20\)
\(\Leftrightarrow\left(x+21\right):8=20-12\)
\(\Leftrightarrow\left(x+21\right):8=8\)
\(\Leftrightarrow x+21=64\)
\(\Leftrightarrow x=43\)
C) \(\frac{1}{2}+\frac{3}{4}x=\frac{3}{2}\)
\(\Leftrightarrow\frac{3}{4}x=\frac{3}{2}-\frac{1}{2}\)
\(\Leftrightarrow\frac{3}{4}x=1\)
\(\Leftrightarrow x=\frac{4}{3}\)
\(1,\Leftrightarrow7x=42\Leftrightarrow x=6\\ 2,\Leftrightarrow36-4x=4\Leftrightarrow4x=32\Leftrightarrow x=8\\ 3,\Leftrightarrow3x=35\Leftrightarrow x=\dfrac{35}{3}\\ 4,\Leftrightarrow x-12=144\Leftrightarrow x=156\\ 5,\Leftrightarrow x-14=16\Leftrightarrow x=30\\ 6,\Leftrightarrow3x-24=\dfrac{148}{73}\Leftrightarrow3x=\dfrac{1900}{73}\Leftrightarrow x=\dfrac{1900}{219}\\ 7,\Leftrightarrow33+x=45\Leftrightarrow x=12\\ 8,Sai.đề\\ 9,\Leftrightarrow\left(x+9\right):2=39\Leftrightarrow x+9=78\Leftrightarrow x=69\\ 11,\Leftrightarrow2\left(x+7\right)=38\Leftrightarrow x+7=19\Leftrightarrow x=12\\ 13,\Leftrightarrow2\left(x-51\right)=66\Leftrightarrow x-51=33\Leftrightarrow x=84\)
\(15,\Leftrightarrow x-19=9\Leftrightarrow x=28\\ 17,\Leftrightarrow\left(x-3\right):2=48\Leftrightarrow x-3=96\Leftrightarrow x=99\\ 19,\Leftrightarrow0x=46\Leftrightarrow x\in\varnothing\\ 8,Sai.đề\\ 14,\Leftrightarrow2x=209\Leftrightarrow x=\dfrac{209}{2}\\ 16,\Leftrightarrow2x+6=157\Leftrightarrow2x=151\Leftrightarrow x=\dfrac{151}{2}\\ 18,\Leftrightarrow5\left(x+4\right)=100\Leftrightarrow x+4=20\Leftrightarrow x=16\\ 20,\Leftrightarrow\left(3x-5\right)^3=27=3^3\Leftrightarrow3x-5=3\Leftrightarrow x=\dfrac{8}{3}\)
a)\(\left(5x+1\right)^2=\frac{36}{49}\\ \left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{matrix}\right.\)
vậy...
2.
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
⇒ \(5x+1=\pm\frac{6}{7}\)
⇒ \(\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x=\frac{6}{7}-1=-\frac{1}{7}\\5x=\left(-\frac{6}{7}\right)-1=-\frac{13}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-\frac{1}{7}\right):5\\x=\left(-\frac{13}{7}\right):5\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{35};-\frac{13}{35}\right\}.\)
Chúc bạn học tốt!
Bài 1:
a. Ta có: $1953=3^2.7.31$
$777=3.7.37$
$\Rightarrow UCLN(1953, 777) = 3.7=21$
b. Đề khó hiểu quá. Bạn xem lại/
Bài 2: Thiếu dấu giữa các số. Bạn xem lại.