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a, \(\left(x+2\right)^2-5=4\Rightarrow\left(x+2\right)^2=5+4\)
\(\Rightarrow\left(x+2\right)^2=9=3^2\Rightarrow x+2=3\Rightarrow x=3-2=1\)
Vậy x = 1
b, \(\left|1-x\right|+2=-1\Rightarrow\left|1-x\right|=-1+\left(-2\right)\)
\(\Rightarrow\left|1-x\right|=-3\Rightarrow x\in\varnothing\)
Vậy \(x\in\varnothing\)
c, \(x^2=4x\Leftrightarrow x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\)
\(\Rightarrow\left[\begin{matrix}x=0\\x-4=0\end{matrix}\right.\) \(\Leftrightarrow\left[\begin{matrix}x=0\\x=0+4=4\end{matrix}\right.\)
Vậy x = 0 hoặc x = 4
a) \(\left(x+2\right)^2-5=4\)
\(\Rightarrow\left(x+2\right)^2=9\)
\(\Rightarrow x+2=3\) hoặc x + 2 = -3
+) \(x+2=3\Rightarrow x=1\)
+) \(x+2=-3\Rightarrow x=-5\)
Vậy \(x\in\left\{1;-5\right\}\)
b) \(\left|1-x\right|+2=-1\)
\(\Rightarrow\left|1-x\right|=-3\)
Mà \(\left|1-x\right|\ge0\)
\(\Rightarrow x\) không có giá trị thỏa mãn
Vậy x không có giá trị thỏa mãn
c) \(x^2=4x\)
\(\Rightarrow x^2-4x=4x-4x\)
\(\Rightarrow x^2-4x=0\)
\(\Rightarrow x\left(x-4\right)=0\)
\(\Rightarrow\left[\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy \(x\in\left\{0;4\right\}\)
a,\(\frac{4}{9}.\frac{2}{6}=\frac{4}{27}\)
b,\(1\frac{1}{3}.\left(0,5\right)+\left(\frac{8}{15}-\frac{19}{30}\right):\frac{6}{15}\)
=\(\frac{4}{3}.\frac{1}{2}+\left(\frac{16}{30}-\frac{19}{30}\right).\frac{15}{6}\)
=\(\frac{2}{3}+\frac{-1}{10}.\frac{15}{6}\)
=\(\frac{2}{3}+\frac{-1}{4}\)
=\(\frac{8}{12}+\frac{-3}{12}=\frac{5}{12}\)
bài2
a,\(\left(\frac{2}{7}.x+\frac{3}{7}\right):2\frac{1}{5}-\frac{3}{7}=1\)
=>\(\left(\frac{2}{7}.x+\frac{3}{7}\right):\frac{11}{5}=1+\frac{3}{7}=\frac{10}{7}\)
=>\(\frac{2}{7}.x+\frac{3}{7}=\frac{10}{7}.\frac{11}{5}\)
=>\(\frac{2}{7}.x+\frac{3}{7}=\frac{22}{7}\)
=>\(\frac{2}{7}.x=\frac{22}{7}-\frac{3}{7}=\frac{19}{7}\)
=>\(x=\frac{19}{7}:\frac{2}{7}=\frac{19}{7}.\frac{7}{2}=\frac{19}{2}\)
vậy x\(=\frac{19}{2}\)
Bài 1:
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c: \(\Leftrightarrow\left|x-1\right|-1=1\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
Bài 2:
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
Bài 3:
a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)
Dấu '=' xảy ra khi x=-15/19
b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=4/7
Bài 1:
\(a,22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)
=\(\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)
=\(\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)
=\(\frac{70}{4}+\frac{2}{4}-\frac{5}{4}\)
=\(\frac{67}{4}\)
\(b,1,4.\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{12}{15}+\frac{10}{15}\right):\frac{11}{5}\)
=\(\frac{3}{7}-\frac{22}{15}.\frac{5}{11}\)
=\(\frac{3}{7}-\frac{2}{3}\)
=\(-\frac{5}{21}\)
\(c,125\%.\left(-\frac{1}{2}\right)^2:\left(1\frac{5}{6}-1,6\right)+2016^0\)
=\(\frac{5}{4}.\frac{1}{4}:\left(\frac{11}{6}-\frac{8}{5}\right)+1\)
=\(\frac{5}{16}:\frac{7}{30}+1\)
=\(\frac{131}{56}\)
\(d,1,4.\frac{15}{49}-\left(20\%+\frac{2}{3}\right):2\frac{1}{5}\)
=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{1}{5}+\frac{2}{3}\right):\frac{11}{5}\)
=\(\frac{3}{7}-\frac{13}{15}:\frac{11}{5}\)
=\(\frac{3}{7}-\frac{13}{33}\)
=\(\frac{8}{231}\)
Bài đ làm giống hệt như bài c
Bài 2 :
\(a,\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)
=>\(\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}=1\\x=\frac{1}{4}:\frac{3}{4}=\frac{1}{3}\end{matrix}\right.\)
Vậy x ∈{1;\(\frac{1}{3}\)}
\(b,\frac{5}{3}.x-\frac{2}{5}.x=\frac{19}{10}\)
=>\(\frac{19}{15}.x=\frac{19}{10}\)
=>\(x=\frac{19}{10}:\frac{19}{15}=\frac{3}{2}\)
Vậy x ∈ {\(\frac{3}{2}\)}
c,\(\left|2.x-\frac{1}{3}\right|=\frac{2}{9}\)
=>\(\left[{}\begin{matrix}2.x-\frac{1}{3}=\frac{2}{9}\\2.x-\frac{1}{3}=-\frac{2}{9}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2.x=\frac{2}{9}+\frac{1}{3}=\frac{5}{9}\\2.x=-\frac{2}{9}+\frac{1}{3}=\frac{1}{9}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\frac{5}{9}:2=\frac{5}{18}\\x=\frac{1}{9}:2=\frac{1}{18}\end{matrix}\right.\)
Vậy x∈{\(\frac{5}{18};\frac{1}{18}\)}
\(d,x-30\%.x=-1\frac{1}{5}\)
=\(70\%x=-\frac{6}{5}\)
=\(\frac{7}{10}.x=-\frac{6}{5}\)
=>\(x=-\frac{6}{5}:\frac{7}{10}=-\frac{12}{7}\)
Vậy x∈{\(-\frac{12}{7}\)}
Bài 2
a/
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{3}{4}\\\frac{3}{4}.x=\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}\\x=\frac{1}{4}:\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=\frac{1}{3}\)
b/ Đặt x làm thừa số chung rồi tính như bình thường
c/ Tương tự câu a
d/ Tương tự câu b
1, Ta có :
a . 81 = 34 => 3x= 34 => x = 4 .
b. 125 = 53 => 5x+2 = 53 =>x + 2 = 3 => x = 1
c. 23 * 2x - 1 = 64
=> 23 + ( x - 1 ) = 64 = 26
=> 3 + ( x - 1 ) = 6
=> x - 1 = 6 - 3 = 3
x = 3 + 1
x = 4
Tìm x
190−(x−1)2=189
(x - 1) 2 = 190 - 189
(x - 1) 2 = 1
x = 1+2
x = 2
a, cái câu trả lời trước mk ghi sai
190 - (x-1)2 = 189
(x - 1 ) 2 = 190 - 189
(x-1)2 = 1
x = 1+1
x = 2