Bài 1 : Tìm x
a) 2/3 - 1/3 (x-3/2)-1/2(2x+1)=5
b) (x+1/2)(X-3/4)=0
c) 1/3 x + 3/5 (x+1)=0
d)x-8/2 > 0
e) x:(2/9 - 1/5 )=1/2
g) 1/3 + 1/2 : x = 1/5
Bài 1 : Tìm x
a) 2/3 - 1/3 (x-3/2)-1/2(2x+1)=5
b) (x+1/2)(X-3/4)=0
c) 1/3 x + 3/5 (x+1)=0
d)x-8/2 > 0
e) x:(2/9 - 1/5 )=1/2
g) 1/3 + 1/2 : x = 1/5
Nhanh Nha
Tìm x:
a) 2 3/4 - x=3/4
b) x:5/6=-3/5
c)1 1/3 +2/3:x=1
d) x-1/9=8/3
e) 1/2 x + 650%x-x= -6
g) 2(x - 1/2) + 3(-1+x/3)=x(2/x - 1) (x khác 0)
h) x-2/20= -5/2-x
i) (x/2-1)3 + 2=-11/8
k) (x/3 +1/2) (75% - 1 1/2x)=0
GIÚP MÌNH VỚI Ạ. CẢM ƠN MỌI NGƯỜI!
a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)
b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)
\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)
c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)
\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)
\(\Rightarrow x=-2\)
d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)
\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)
\(\Rightarrow x=\dfrac{25}{9}\)
e) \(\dfrac{1}{2}x+650\%x-x=-6\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)
\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)
\(\Rightarrow6x=-6\)
\(\Rightarrow x=\dfrac{-6}{6}=-1\)
g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)
\(\Rightarrow2x-1-3+x=2-x\)
\(\Rightarrow3x-4=2-x\)
\(\Rightarrow3x+x=2+4\)
\(\Rightarrow4x=6\)
\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)
h) \(x-\dfrac{2}{20}=-\dfrac{5}{2}-x\)
\(\Rightarrow x+x=-\dfrac{5}{2}+\dfrac{2}{20}\)
\(\Rightarrow2x=-\dfrac{12}{5}\)
\(\Rightarrow x=-\dfrac{12}{5}:2=-\dfrac{6}{5}\)
i) \(\left(\dfrac{x}{2}-1\right)^3+2=-\dfrac{11}{8}\)
\(\Rightarrow\left(\dfrac{x}{2}-1\right)^3=-\dfrac{11}{8}-2\)
\(\Rightarrow\dfrac{x}{2}-1=\sqrt[3]{-\dfrac{27}{8}}\)
\(\Rightarrow\dfrac{x}{2}-1=-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{x}{2}=-\dfrac{3}{2}+1\)
\(\Rightarrow x=-\dfrac{1}{2}.2=-1\)
k) \(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\\dfrac{3}{4}-1\dfrac{1}{2}x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\\dfrac{3}{2}x=\dfrac{3}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}.3=-\dfrac{3}{2}\\x=\dfrac{3}{4}:\dfrac{3}{2}=\dfrac{1}{2}\end{matrix}\right.\)
a,x+5/x-1+8/x^2-4x+3=x+1/x-3 b,x-4/x-1-x^2+3/1-x^2+5/x+1=0 c,3x/4-5=3-x/2+5x-1/6 d,(x-2)(x+2)-(x-3)(x+4)-2x+3=0 e,(x-1)^2+2(x+1)=5x+5 g,(x-3)(x+4)x=0
a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)
=>(x+5)(x-3)+8=x^2-1
=>x^2+2x-15+8=x^2-1
=>2x-7=-1
=>x=3(loại)
b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)
=>(x-4)(x+1)+x^2+3+5(x-1)=0
=>x^2-3x-4+x^2+3+5x-5=0
=>2x^2+2x-6=0
=>x^2+x-3=0
=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)
e: =>x^2-2x+1+2x+2=5x+5
=>x^2+3=5x+5
=>x^2-5x-2=0
=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)
g: (x-3)(x+4)*x=0
=>x=0 hoặc x-3=0 hoặc x+4=0
=>x=0;x=3;x=-4
Bài 1 : Tìm x
a) 2/3 - 1/3 (x-3/2)-1/2(2x+1)=5
b) (x+1/2)(X-3/4)=0
c) 1/3 x + 3/5 (x+1)=0
d)x-8/2 > 0
e) x:(2/9 - 1/5 )=1/2
g) 1/3 + 1/2 : x = 1/5
a: =>2/3-1/3x+1/2-x-1/2=5
=>-4/3x+2/3=5
=>-4/3x=13/3
=>x=-13/4
b: \(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\x-\dfrac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
c: =>1/3x+3/5x+3/5=0
=>14/15x=-3/5
=>x=-3/5:14/15=-3/5x15/14=-45/70=-9/14
d: =>x>8/2
e: =>x:1/45=1/2
=>x=1/90
g: =>1/2:x=-2/15
=>x=-1/2:2/15=-15/4
Tìm x:
Bài 3:
a) 1/3 + 2/3 : x = -7
b) 3 1/2 - 1/2 x = 2/3
c) [(x + 1/3) * 3/4 +5] : 2=3
d) (1 2/3 - x). 0,75 - 2 = 1/2
e) x + 75% x = -1,6
f) (x - 2/3) (2x+1) = 0
g) 1 - (2x + 1/2) 2 = 3/4
h) 5/1.6 + 5/6.11 + ... + 5/(5x + 1).(5x + 6) = 2020/2021
Giúp mình bài 3 với, ai làm được đúng hết mình sẽ tik hết nhé! Hurry up!
Tìm x biết
a,3/1/2x -5=0
b,3×(x-2)+5/3/4=0
c,1/2x-3-2=0
d, 1/3x+1=2x-3/2
e,(x)-3=5/1/2
g,(x-5)(x+1/2)=0
H,(2x-1/2)=4
i,(4+x3)=8
Bài 1: rút gọn
a. ( x + 1)^2 - (x - 1)^2 - 3(x + 1)(x - 1)
b. 5(x +2)(x -2) - 1/2(6 - 8x)^2 + 17
c. (x^2 - 1)^3 - (x^4 + x^2 + 1)(x^2 - 1)
d. (x^4 - 4x^2 +9)(x^2 + 3) - (3 + x^2)^3
e. (x-3)^3 -(x - 3)(x^2 +3x + 9) + 6(x + 1)^2
Bài 2: tìm x
a. 25x^2 - 9 = 0
b. (x + 4)^2 - (x + 1) (x -1) = 16
c. (2x - 1)^2 + (x + 3)^2 - 5(x + 7)(x - 7) = 0
d. (x + 2)(x^2 - 2x + 4) - x(x^2 + 2)= 15
e. (x + 3)^3 - x(3x + 1)^2 + (2x + 1)(4x^2 - 2x + 1) = 28
g. (x^2 - 1)^3 - (x^4 + x^2 + 1)(x^2 -1) = 0
HELP ME!!!!!!!!
Bn gửi từng câu sẽ có nhều ng trl hơn nhé
tý mk giải câu a cho cần ko
Tìm x
a) 1/3x + 2/5( x - 1 ) = 0
b) (2x - 3 )(6 - 2x ) =0
c) 2|1/2x - 1/3 | - 3/2 = 1/4
d) 3/4 - 2 . | 2x - 2/3 | = 2
e) ( 3x - 1)(-1/2x + 5 ) = 0
g)-5(x + 1/5 ) - 1/2(x - 2/3 ) = 2/3x - 5/6
h) 3(x - 1/2 ) - 5(x + 3/5 ) = -3 + 1/5
i) 60%x + 2/3x = 1/3. 6và1/3 ( 6và1/3 là hỗn số )
Bài 1 tìm x
l) (x + 9) . (x2 – 25) = 0
e) |x - 4 |< 7
f) 40 < 31 + |x |< 47
g) | x + 3| ≤ 2
m) (-5x + 20).(x3 – 8) = 0
a) (x + 1).(y - 2) = 5
b) (x - 5).(y + 4) = -7
c) (x + 1)2 + (y – 1)2 = 0
d) (2x – 18)2 + ( y + 37)2 = 0
k |x-40|+|x-y+10|_<0
l) (x + 9) . (x2 – 25) = 0
<=> (x + 9) . (x – 5) . (x + 5) = 0
<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{-9,5,-5\right\}\)
e) |x - 4 |< 7
<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
Vậy S = \(\left\{11;-3\right\}\)
I,(x+9).(x^2-25)=0
tương đương:x+9=0
x^2-25=0
tương đương : x=-9
x=5
e,\(\left|x-4\right|\)=7
tương đương x-4=4
x-4=-4
tương đương :x=0
x=-8
Bài 1:
l) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{-9;5;-5\right\}\)
e) Ta có: |x-4|<7
mà \(\left|x-4\right|\ge0\forall x\)
nên \(\left|x-4\right|\in\left\{0;1;2;3;4;5;6\right\}\)
\(\Leftrightarrow x-4\in\left\{0;1;-1;2;-2;3;-3;4;-4;5;-5;6;-6\right\}\)
hay \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)
Vậy: \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)
f) Ta có: \(40< 31+\left|x\right|< 47\)
\(\Leftrightarrow\left|x\right|+31\in\left\{41;42;43;44;45;46\right\}\)
\(\Leftrightarrow\left|x\right|\in\left\{10;11;12;13;14;15\right\}\)
hay \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)
Vậy: \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)
g) Ta có: \(\left|x+3\right|\le2\)
\(\Leftrightarrow\left|x+3\right|\in\left\{0;1;2\right\}\)
\(\Leftrightarrow x+3\in\left\{0;1;-1;2;-2\right\}\)
hay \(x\in\left\{-3;-2;-4;-1;-5\right\}\)
Vậy: \(x\in\left\{-3;-2;-4;-1;-5\right\}\)