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d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
\(a,\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{5}{6}-\dfrac{2}{3}\\ \Rightarrow\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{1}{6}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{6}+\dfrac{7}{12}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{3}{4}:\dfrac{3}{4}\\ \Rightarrow x=1\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{5}{7}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{-5}{-7}\\ \Rightarrow x=-7\\ c,2\dfrac{1}{3}:x=7\\ \Rightarrow\dfrac{7}{3}:x=7\\ \Rightarrow x=\dfrac{7}{3}:7\\ \Rightarrow x=\dfrac{1}{3}\)
\(d,\dfrac{-105}{12}< x< \dfrac{20}{7}\Rightarrow x\in\left\{-8;-7;...;2\right\}\)
a: \(\Leftrightarrow x\cdot\dfrac{3}{4}=\dfrac{3}{4}\)
hay x=1
b: \(\Leftrightarrow x=\dfrac{-28\cdot5}{20}=-7\)
c: \(\Leftrightarrow x=\dfrac{7}{3}:7=\dfrac{1}{3}\)
d: \(\Leftrightarrow-8< x< 3\)
hay \(x\in\left\{-7;-6;-5;-4;-3;-2;-1;0;1;2\right\}\)
B3 a) x=4 b) x=-7 c) x=5 d) x=4
B2 a) -3+ -2+ -1+0+1+2+3+4=4
b) -6+ -5+ -4+ -3+ -2+ -1+0+1+2+3+4=-11
c) -18+-17+-16+-15+-14+-13+-12+-11+-10+-9+-8+-7+-6+-5+-4+3+-2+-1+0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19=19
\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}=\dfrac{28:7}{49:7}=\dfrac{4}{9}\\ Vậy:x=\dfrac{4.9}{4}=9\\ y=\dfrac{4.21}{9}=\dfrac{28}{3}\)
\(\dfrac{x}{2}=\dfrac{3}{y}\\ \Leftrightarrow x.y=2.3=6\\ Vậy:\left[{}\begin{matrix}\left(x;y\right)=\left(1;6\right)=\left(6;1\right)\\\left(x;y\right)=\left(2;3\right)=\left(3;2\right)\end{matrix}\right.\)
[(46-32)^2-(54-42)^2]*36-1872
=(14^2-12^2)*36-1872
=(196-144).36-1872
=52.36-1872
=1872-1872
=0
(x-7)^7=(x-7)^8
=>(x-7)^7-(x-7)^8=0
=>(x-7)^7.1-(x-7)^7.(x-7)=0
=>(x-7)^7.[1-(x-7)]=0
\(\Rightarrow\orbr{\begin{cases}x-7=0\\1-\left(x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}x=7;-7\\x=0\end{cases}}}\)
Vậy x={7;-7;0}
1+2^3+3^3-4^2.x=20
=>36-4^2.x=20
=>4^2.x=16=4^2
=>2.x=2
=>x=1
Bài 1:
a. $(-20)+x=-30$
$x-20=-30$
$x=-30+20=-(30-20)=-10$
b.
$(-10)-x=-20$
$x=(-10)-(-20)=-10+20=20-10=10$
c. Đề sai. Bạn xem lại.
d.
$x+(-3)=-7$
$x=-7-(-3)=-7+3=-(7-3)=-4$
e.
$x-(-5)=-9$
$x=(-9)+(-5)=-14$
f.
$x(-11)=12$
$x=\frac{12}{-11}=\frac{-12}{11}$
h.
$2x-10=20$
$2x=20+10=30$
$x=30:2=15$
l.
$4x-8=-8$
$4x=-8+8=0$
$x=0:4=0$
k.
$-12-(-2)x=-8$
$(-2)x=-12-(-8)=-12+8=-(12-8)=-4$
$x=(-4):(-2)=2$
Bài 2:
a. $-20-(10-x)=-3$
$10-x=-20-(-3)=-20+3=-(20-3)=-17$
$x=10-(-17)=10+17=27$
b.
$14+(14-x)=-2$
$14-x=-2-14=-16$
$x=14-(-16)=14+16=30$
c.
$-15-(x-3)=-7$
$x-3=-15-(-7)=-15+7=-8$
x=-8+3=-5$
d.
$(x+4)+(-20)=-8$
$x+4=-8-(-20)=-8+20=12$
$x=12-4=8$
e.
$-2x-2=-4$
$-2x=-4+2=-2$
$x=(-2):(-2)=1$
f.
$-2x+4=-4$
$-2x=-4-4=-8$
$x=(-8):(-2)=4$
l.
$-12-(-2)x=-2-4=-6$
$(-2)x=-12-(-6)=-12+6=-6$
$x=(-6):(-2)=3$
a) \(\frac{x^7}{3^5}=9\)
=> \(\frac{x^7}{3^5}=3^2\)
=> x7 = 32 . 35 = 37
=> x = 3
b) \(\frac{32}{x^7}=4\)
=> \(\frac{2^5}{x^7}=2^2\)
=> \(2^2\cdot x^7=2^5\)
=> \(x^7=\frac{2^5}{2^2}=2^3\)
=> không tìm được x
2. \(\frac{2^{30}}{3^{20}}-\left(\frac{2}{3}\right)^{20}\cdot2^{10}\)
\(=\frac{2^{30}}{3^{20}}-\frac{2^{20}}{3^{20}}\cdot2^{10}\)
\(=\frac{2^{30}}{3^{20}}-\frac{2^{30}}{3^{20}}=0\)