[124-(20+4x)]:30+7=11
Ai nhanh mik k cho
nhớ giải đầy đủ nha
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\(\dfrac{11}{3}-\dfrac{3}{11}=\dfrac{121}{33}-\dfrac{99}{33}=\dfrac{121-99}{33}=\dfrac{22}{33}\)
a) 7^2.5-(2x+1)=630:9
b) (10-4x)+120:2^3=17
c) [124-(20-4X)]:30+7=11
d) |2x-5|=1
nhớ ghi đầy đủ bước nhá
\(a\)\(.\)\(7^2.5-\left(2x+1\right)=630:9\)
\(49.5-\left(2x+1\right)=70\)
\(245-\left(2x+1\right)=70\)
\(2x+1=245-70\)
\(2x+1=175\)
\(2x=175-1\)
\(2x=174\)
\(x=174:2\)
\(x=87\)
Vậy \(x=87\)
\(b.\)\(\left(10-4x\right)+120:2^3=17\)
\(\left(10-4x\right)+120:8=17\)
\(\left(10-4x\right)+15=17\)
\(10-4x=17-15\)
\(10-4x=2\)
\(4x=10-2\)
\(4x=8\)
\(x=8:4\)
\(x=2\)
Vậy \(x=2\)
\(c.\)\(\left[124-\left(20-4x\right)\right]:30+7=11\)
\(\left[124-\left(20-4x\right)\right]:30=11-7\)
\(\left[124-\left(20-4x\right)\right]:30=4\)
\(\left[124-\left(20-4x\right)\right]=4.30\)
\(\left[124-\left(20-4x\right)\right]=120\)
\(124-20+4x=120\)
\(104+4x=120\)
\(4x=120-104\)
\(4x=16\)
\(x=16:4\)
\(x=4\)
Vậy \(x=4\)
\(d.\)\(\left|2x+5\right|=1\)
\(\Rightarrow2x+5\in\left\{-1;1\right\}\)
\(\Rightarrow2x\in\left\{-6;-4\right\}\)
\(\Rightarrow x\in\left\{-3;-2\right\}\)
Vậy \(x\in\left\{-3;-2\right\}\)
học tốt
a) \(7^2\cdot5-\left(2x+1\right)=630:9\)
\(\Leftrightarrow49\cdot5-2x-1=70\)
\(\Leftrightarrow244-2x=70\)
\(\Leftrightarrow2x=244-70\)
\(\Leftrightarrow2x=174\)
\(\Leftrightarrow x=87\)
Vậy x=87.
b) \(\left(10-4x\right)+120:2^3=17\)
\(\Leftrightarrow10-4x+120:8=17\)
\(\Leftrightarrow10-4x+15=17\)
\(\Leftrightarrow25-4x=17\)
\(\Leftrightarrow4x=25-17=8\)
\(\Leftrightarrow x=2\)
Vậy x=2
c)\([124-\left(20-4x\right)]:30+7=11\)
\(\Leftrightarrow\left(124-20+4x\right):30=4\)
\(\Leftrightarrow104+4x=120\)
\(\Leftrightarrow4x=120-104=16\)
\(\Leftrightarrow x=4\)
Vậy x=4
d) \(|2x-5|=1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=1\\2x-5=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
Vậy \(x\in\text{ }\left\{3;2\right\}\)
Bổ sung đề bài : Tính
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}=\frac{6}{7}\)
\(\sqrt{4x^2-4x+9}=3\\ \Rightarrow4x^2-4x+9=9\\ \Rightarrow4x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Ta có: \(\sqrt{4x^2-4x+9}=3\)
\(\Leftrightarrow4x^2-4x=0\)
\(\Leftrightarrow4x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Ta có: 2^30= 2^3.10= (2^3)^10= 8^10
3^20= 3^2.10= (3^2)^10= 9^10
Vì 8<9 nên 8^10<9^10
=> 2^30<3^20
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{6.7}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
Lời giải:
$1+2+3+....+25$
$=(1+25)+(2+24)+(3+23)+(4+22)+(5+21)+(6+20)+(7+19)+(8+18)+(9+17)+(10+16)+(11+15)+(12+14)+13$
$=\underbrace{26+26+26+...+26}_{12}+13$
$=26\times 12+13=325$
[124-(20+4x)]:30+7=11
[124-(20+4x)]:30 = 11-7
[124-(20+4x)]:30 = 4
[124-(20+4x)] =4 x 30
[124-(20+4x)] =120
[124-20-4x] =120
104-4x = 120
-4x = 120-104
-4x = 16
x = -4
20 + 4 x ? hả cậu ?