Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
( 2x + 3 ) 3 = -125
( 2x + 3 ) 3 = ( -5 ) 3
=> 2x + 3 = -5
2 x = -8
x = -4
( x + 1 / 4 ) 4 = 625 / 256
( x + 1 / 4 ) 4 = ( 5 / 4 ) 4
=> x + 1 / 4 = 5/4
x = 4 / 4
x = 1
Bài 5:
A 1 2 3 4 B 1 C 1 D 1
Ta có : \(\widehat{A_1}+\widehat{A_3}=180^o\) (kề bù)
\(100^o+\widehat{A_3}=180^o\)
\(\widehat{A_3}=80^o\)
Ta có: \(\widehat{A_3}=\widehat{B_1}=80^o\)
\(\widehat{A_3}\) và \(\widehat{B_1}\) ở vị trí đồng vị
\(\Rightarrow AC//BD\)
\(\Rightarrow\widehat{C}_1=\widehat{D_1}=135^o\) (đồng vị)
\(x=135^o\)
b)
G H B K 1 1 1 1
Ta có: \(\widehat{G_1}+\widehat{B_1}=180^o\left(120^o+60^o=180^o\right)\)
\(\widehat{G_1}\) và \(\widehat{B_1}\) ở vị trí trong cùng phía
\(\Rightarrow QH//BK\)
\(\Rightarrow\widehat{H_1}=\widehat{K_1}=90^o\)(so le)
\(x=90^o\)
bạn giúp mình bài 7 trang 10 toán 7 ki 1 nhé
bạn có sdt ko cho mình xin với
|2x - 3| + x = 2
=> |2x - 3| = 2 - x
+ Với \(x< \frac{3}{2}\) thì |2x - 3| = 3 - 2x
Ta có: 3 - 2x = 2 - x
=> 3 - 2 = -x + 2x
=> x = 1, thỏa mãn \(x< \frac{3}{2}\)
+ Với \(x\ge\frac{3}{2}\) thì |2x - 3| = 2x - 3
Ta có: 2x - 3 = 2 - x
=> 2x + x = 2 + 3
=> 3x = 5
=> \(x=\frac{5}{3}\), thỏa mãn \(x\ge\frac{3}{2}\)
Vậy \(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{3}\end{array}\right.\)
\(\Rightarrow\dfrac{3}{4}\cdot\dfrac{9}{22}-\left|-3x+\dfrac{8}{3}\right|=\dfrac{3}{4}\\ \Rightarrow\left|-3x+\dfrac{8}{3}\right|=\dfrac{11}{6}-\dfrac{3}{4}=\dfrac{13}{12}\\ \Rightarrow\left[{}\begin{matrix}-3x+\dfrac{8}{3}=\dfrac{13}{12}\\3x-\dfrac{8}{3}=\dfrac{13}{12}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=\dfrac{19}{12}\\3x=\dfrac{15}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{19}{36}\\x=\dfrac{5}{4}\end{matrix}\right.\)
\(\dfrac{3}{4}:2\dfrac{4}{9}-\left|-3x+2\dfrac{2}{3}\right|=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{3}{4}:\dfrac{22}{9}-\left|-3x+\dfrac{8}{3}\right|=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{27}{88}-\left|-3x+\dfrac{8}{3}\right|=\dfrac{3}{4}\)
\(\Rightarrow\left|-3x+\dfrac{8}{3}\right|=-\dfrac{39}{88}\left(VLý\right)\)
Vậy \(S=\varnothing\)
\(a,\frac{5}{6}-2\sqrt{\frac{4}{9}}+\sqrt{\left(-2\right)^2}\)
\(=\frac{5}{6}-2.\frac{2}{3}+2\)
\(=\frac{5}{6}-\frac{4}{6}+\frac{12}{6}\)
\(=\frac{5-4+12}{6}=\frac{13}{6}\)
\(b,\left(-3\right)^2.\left(\frac{1}{3}\right)^3:\left[\left(-\frac{2}{3}\right)^3-1\frac{1}{3}\right]-\left(-200\right)^0\)
\(=9.\frac{1}{27}:\left(-\frac{8}{27}-\frac{5}{3}\right)-1\)
\(=\frac{1}{3}:\left(-\frac{8}{27}-\frac{45}{27}\right)-1\)
\(=\frac{1}{3}:\left(-\frac{53}{27}\right)-1\)
\(=\frac{1}{3}.\left(-\frac{27}{53}\right)-1\)
\(=-\frac{9}{53}-1=-\frac{9}{53}-\frac{53}{53}\)
\(=-\frac{62}{53}\)
\(c,\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):2\)
\(=\left(-\frac{1}{2}-\frac{3}{5}\right).\frac{1}{3}+\frac{1}{3}-\left(-\frac{1}{6}\right).\left(-\frac{1}{2}\right)\)
\(=\left(-\frac{5}{10}-\frac{6}{10}\right).\frac{1}{3}+\frac{1}{3}-\frac{1}{12}\)
\(=-\frac{11}{10}.\frac{1}{3}+\frac{1}{3}-\frac{1}{12}\)
\(=\frac{1}{3}\left(-\frac{11}{10}-\frac{1}{12}\right)\)
\(=\frac{1}{3}\left(-\frac{66}{60}-\frac{5}{60}\right)\)
\(=\frac{1}{3}.\left(-\frac{71}{60}\right)\)
\(=-\frac{71}{180}\)
câu
nào mình giúp cho nha 22 uk