Có cái gì đó sai sai

x=0,1976423538

tớ tính bằng máy tính casio đó !(mình là hs lớp 6 mà)

a: =>x+41=7/4

hay x=-157/4

a) 4(x+41)=7
x+41=7/4
x=7/4-41
x=157/4
b) có gì đó sai sai-.-

\(2\left(x-51\right)-22=20.\)

\(2\left(x-51\right)=20+22.\)

\(2\left(x-51\right)=42.\)

\(x-51=42:2.\)

\(x-51=21.\)

\(x=21+51.\)

\(x=72.\)

Vậy.....

~ Học tốt!!! ~

\(2(x-51)-22=20\)

\(2x-102=20+22\)

\(2x-102=42\)

\(2x=144\)

\(x=72\)

A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
 

 C = 2^-3 + (5²)³.5^-3 + 4^-3.16 - 2.3² - 105.(\(\dfrac{24}{51}\))⁰

C =  \(\dfrac{1}{8}\) + 5⁶.5^-3 + 4^-3.4² - 2.9 - 105.1

C =  \(\dfrac{1}{8}\) + 5³ + \(\dfrac{1}{4}\) - 18 - 105

C =  (\(\dfrac{1}{8}\) + \(\dfrac{1}{4}\))  - (105 - 125 + 18)

C = \(\dfrac{3}{8}\) - (-20 + 18)

C = \(\dfrac{3}{8}\)  + 2

C = \(\dfrac{19}{8}\)

D = 2^-3 . \(\dfrac{1}{2^{-2}}\).\(\dfrac{2}{3}\) + 4^-2.8 - 7. (\(\dfrac{17}{23}\))⁰ + 19

D = 2^-1.\(\dfrac{2}{3}\) + \(\dfrac{1}{16}\).8 - 7.1 + 19

D = \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\) - 7 + 19

D = \(\dfrac{5}{6}\)  + (19 - 7)

D = \(\dfrac{5}{6}\) + 12

D = \(\dfrac{77}{6}\)

\(2\left(x-51\right)=2.2^3+20\)

\(\Leftrightarrow2\left(x-51\right)=16+20\)

\(\Leftrightarrow2\left(x-51\right)=36\)

\(\Leftrightarrow x-51=13\)

\(\Leftrightarrow x=64\)

Vậy ..

cho mk hỏi xíu 2,2\(^3\) + 20 hay là 2\(\times\)2\(^3\)+20 vậy

2(x-51)=2.2³+20

2(x-51)=2⁴+20

x-51=36:2

x=18+51

x=69

5,6 * '' 2,8 + 2 + 8 + 4 ''

5,6 * 16,8

94,08

k tớ nhé

Nhân hết trước, rồi cộng:

5.6 x 2 + 2.8 x 8 + 5.6 x 4

=11.2 + 22.4 + 22.4

=56

hoặc tính ra từng bước.

5.6+[2+8]*4=45.6

a) 3^{x + 2} + 5.3^{x + 1} = 648

<=> 3^{x + 1}.8 = 648

<=> 3^{x + 1} = 81

<=> 3^{x + 1} = 3⁴ 

<=> x + 1 = 4

<=> x = 3

Sửa lại đề: b) 2^{2x + 1}  + 2^{2x + 4} = 3².8⁴

<=> 2^{2x + 1}.9 = 3² . 8⁴ 

<=> 2^{2x + 1} = 8⁴ 

<=> 2^{2x + 1}  = 2¹² 

<=> 2x + 1 = 12 

<=> \(x=\dfrac{11}{2}\)

a) \(|2x|.|3.5|=|-2.8|\)

\(\Rightarrow|2x|.|15|=|-16|\)

\(\Rightarrow|2x|.15=16\)

\(\Rightarrow|2x|=\dfrac{16}{15}\)

Ta có hai trường hợp:

1) \(2x\ge0\)

\(\Rightarrow|2x|=2x=\dfrac{16}{15}\Rightarrow x=\dfrac{8}{15}\)

2) \(2x< 0\)

\(\Rightarrow|2x|=-2x=\dfrac{16}{15}\Rightarrow-x=\dfrac{8}{15}\Rightarrow x=-\dfrac{8}{15}\)

Vậy \(x=\dfrac{8}{15}\) hoặc \(x=-\dfrac{8}{15}\).

b) \(|x-\dfrac{1}{2}|+|x+y|=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\x+y=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)