a) 34-x=27
b) (x-4/5)2=0
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a) \(x+\dfrac{4}{9}=\dfrac{5}{27}\)
\(x=\dfrac{5}{27}-\dfrac{4}{9}\)
\(x=-\dfrac{7}{27}\)
b) \(x-\dfrac{4}{11}=\dfrac{7}{33}\)
\(x=\dfrac{7}{33}+\dfrac{4}{11}\)
\(x=\dfrac{19}{33}\)
c) \(\dfrac{8}{5}-x=\dfrac{1}{3}\times\dfrac{2}{5}\)
\(\dfrac{8}{5}-x=\dfrac{2}{15}\)
\(x=\dfrac{8}{5}-\dfrac{2}{15}\)
\(x=\dfrac{22}{15}\)
d) \(x-\dfrac{3}{4}=\dfrac{1}{2}+\dfrac{2}{6}\)
\(x-\dfrac{3}{4}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}+\dfrac{3}{4}\)
\(z=\dfrac{19}{12}\)
a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)^2=27.\)
\(\Leftrightarrow x^3+27-x\left(x^2-4x+4\right)-27=0.\)
\(\Leftrightarrow x^3-x^3+4x^2-4x=0.\)
\(\Leftrightarrow4x\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0.\\x-1=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0.\\x=1.\end{matrix}\right.\)
Vậy \(S=\left\{0;1\right\}.\)
Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
5+5+8+0+6+4+5+2+4+1+1+2+3+4+5+6+6+7+8+9+100000000+45638+78536 x 12345 x 34 x 0 +100=100045829
CHÚC BẠN HỌC GIỎI
TK MÌNH NHÉ
Dạ, cảm ơn anh, nhưng đây kg phải là đáp án của em. TRẻ con còn biết, mà người lại kg .Anh đừng tự tin, nhanh rồi cũng có lúc sai!!!
a: \(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)
\(\Leftrightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)
hay \(x=\dfrac{5}{6}\)
\(a,\left(x-12\right)\times1000=0\)
\(x-12=0\)
\(x=0+12\)
\(x=12\)
\(b,\left(23-x\right)\times34=34\)
\(23-x=34:34\)
\(23-x=1\)
\(x=23-1\)
\(x=22\)
\(c,\left(x-5\right)\times6=24\)
\(x-5=24:6\)
\(x-5=4\)
\(x=4+5\)
\(x=9\)
\(d,2x+3=15\)
\(2x=15-3\)
\(2x=12\)
\(x=12:2\)
\(x=6\)
\(e,6\left(7x+1\right)=48\)
\(7x+1=48:6\)
\(7x+1=8\)
\(7x=7\)
\(x=1\)
\(g,\left(x-6\right)\left(x-34\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-6=0\\x-34=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=34\end{cases}}\)
\(h,\left(x-4\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}\)
Bài làm
a) \(\left(x-12\right).1000=0\)
Vì 1000 > 0 \(\Rightarrow x-12=0\Rightarrow x=12\)
b) \(\left(23-x\right).34=34\)
\(\Rightarrow23-x=1\Rightarrow x=22\)
c) \(\left(x-5\right).6=24\Rightarrow x-5=4\Rightarrow x=9\)
d) \(2x+3=15\Rightarrow2x=12\Rightarrow x=6\)
e) \(6\left(7x+1\right)=48\Rightarrow7x+1=8\)
\(\Rightarrow7x=7\Rightarrow x=1\)
g) \(\left(x-6\right).\left(x-34\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-6=0\\x-34=0\end{cases}\Rightarrow\orbr{\begin{cases}x=6\\x=34\end{cases}}}\)
Vậy x= 6 hoặc x= 34
h)\(\left(x-4\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)
Vậy x=4 hoặc x= -2
i) \(x\left(x+1\right).\left(x+2\right)=3\)
..........
Cậu có thể tam khảo bài làm trên đây ạ, học tốt nha ^^
a) \(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b) \(25\left(x-4\right)=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) 5.(x-7)=0⇔x-7=0⇔x=7
b) 25(x-4)=0⇔x-4=0⇔x=4
c) (34-2x).(2x-6)=0
⇔ 34-2x=0 hoặc 2x-6=0
⇔2x=34 hoặc 2x=6
⇔ x=17 hoặc x=3
d) (2019-x).(3x-12)=0
⇔ 2019-x=0 hoặc 3x-12=0
⇔ x=2019 hoặc x=4
e) 57.(9x-27)=0
⇔ 9x-27=0
⇔ x=3
f) 25+(15-x)=30
⇔ 15-x=5
⇔ x=10
g) 43-(24-x)=20
⇔ 24-x=23
⇔ x=1
h) 2.(x-5)-17=25
⇔ 2(x-5)=42
⇔x-5=21
⇔ x=26
i) 3(x+7)-15=27
⇔ 3(x+7)=42
⇔ x+7=14
⇔ x=7
j) 15+4(x-2)=95
⇔ 4(x-2)=80
⇔ x-2=20
⇔ x=22
k) 20-(x+14)=5
⇔ x+14=15
⇔ x=1
l) 14+3(5-x)=27
⇔ 3(5-x)=13
⇔ 5-x=13/3
⇔ x=5-13/3
⇔ x=2/3
a) 34-x = 33
\(\Rightarrow\)4 - x = 3
x = 4 - 3
x = 1
b) ( x - \(\frac{4}{5}\))2 = 02
\(\Rightarrow\)x - \(\frac{4}{5}=0\)
x = 0 + \(\frac{4}{5}\)
x = \(\frac{4}{5}\)
a, đề bài
---> \(3^{4-x}=3^3\)
---> 4 - x = 3
---> x = 1
b, đề bài
---> \(x-\frac{4}{5}=0\)
---> \(x=\frac{4}{5}\)
xin tiick