Tìm giúp mình thằng con tên x với :))
a, 30 . (x+2) - 6. (x-5) - 24 . x = 100
b, x . ( x+3) = 0
c, (x-2) . (5-x) = 0
d, (x-1) . ( x^2+3) = 0
Ai giúp dc mk thì giúp với !!!
câu này 4(x 2) x^2 2x=0
là (x-2) hay (x+2) a
Bạn vào biểu tượng đầu tiên trên thanh công cụ để ghi công thức rõ hơn nhé
Bài 1: tìm x b) x-1,4/1,4-x =2 c) /x-1/ + (2-3x) < 0 d) /-1/2x + 3 / <= 5
Bài 2: tìm x a) /x-1/ + (2-3x) < 0
b) /x-1/ = (-x) - 5
c) /x-2/ + /x-4/ =5
d) /x-1/ + /x-3/ =6
e) / x-1/ + (x+3)
*Các bạn giúp mk với, Thanks nhiều ạ. Các bạn làm đầy đủ thì mình cảm ơn lắm!!!
Bài 1: tìm x b) x-1,4/1,4-x =2 c) /x-1/ + (2-3x) < 0 d) /-1/2x + 3 / <= 5
Bài 2: tìm x a) /x-1/ + (2-3x) < 0
b) /x-1/ = (-x) - 5
c) /x-2/ + /x-4/ =5
d) /x-1/ + /x-3/ =6
e) / x-1/ + (x+3)
*Các bạn giúp mk với, Thanks nhiều ạ. Các bạn làm đầy đủ thì mình cảm ơn lắm!!!
Bài 1:Tìm x
a) ( x - 2 ) . (5 - x) = 0
b) (x - 1) . (x mũ 2 + 1) = 0
c) -12 . (x - 5) + 7 . (3 - x) = 5
d) 30 . (x + 2) - 6 . (x - 5) = 24 . x = 100
Bài 2: Tìm giá trị nhỏ nhất của biểu thức
a) 3 . (x + 1)mũ 2 - 2
b) /x - 2/ - 10
Làm ơn giúp mk vs nè!!~~
1/ a) TH1: x-2 = 0 => x= 0+2 = 2
TH2: 5-x= 0 => x= 5-0 = 5
b)???
duyệt đi
Tìm x:
C, X^2-9=2×(x+3)^2
b, x^3-3x^2+3x-1=0
d, x^2-8x+3x-24=0
Giúp mk với. Mk cảm ơn
c) \(x^2-9=2\cdot\left(x+3\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left[x-3-2\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3-2x-6\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
d) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x^2-8x\right)+\left(3x-24\right)=0\)
\(\Leftrightarrow x\left(x-8\right)+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)
a) \(x^2-9=2\left(x+3\right)^2\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)=2\left(x+3\right)^2\)
\(\Leftrightarrow2\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[2\left(x+3\right)-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[2x+6-x+3\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+9\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+9=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x-8\right)x+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
c) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
1,Tìm X thuộc Z sao cho
(x-7).(x+3)<0
2,tìm x
a,-12.(x-5)-7.(3-x)=5
b,30.(x-2)-6.(x-5)-24.x=100
cả nhà giúp mình vs
bài 9:tìm x
1) (x-3)^2-4=0
2) x^2-2x=24
3) (2x-1)^2+(x+3)^3-5(x+7)(x-7)=0
giúp mình với mn ơi
1) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
2) \(x^2-2x=24\)
\(\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow x^2+4x-6x-24=0\)
\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
Tìm x biết
1) 30.[x+2+6.(x-5)] -24.x=102
2)(x-5).(x^2+1)=0
3) (x+1)+(x+2)+...+(x+99)=0
Các bn làm giúp mk nhanh lên nha . Chiều 2h30 mk đi học rùi 😣😣😣😣
2) \(\left(x-5\right)\left(x^2+1\right)=0\)
\(\Rightarrow x-5=0\) vì \(x^2+1>0\)
\(\Rightarrow x=5\)
cAU 1 TƯƠNG TỰ NHÉ
Tương tự sao được
a/ \(30.\left\{x+2+6\left(x-5\right)\right\}-24x=102\)
\(\Leftrightarrow30.\left\{x+2+6x-30\right\}-24x=102\)
\(\Leftrightarrow30.\left\{7x-28\right\}-24x=102\)
\(\Leftrightarrow210x-340-24x=102\)
\(\Leftrightarrow186x-340=102\)
\(\Leftrightarrow186x=442\)
\(\Leftrightarrow x=\frac{442}{186}\)
c/ \(\left(x+1\right)+\left(x+2\right)+....+\left(x+99\right)=0\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+......+99\right)=0\)
\(\Leftrightarrow99x+49500=0\)
\(\Leftrightarrow x=-50\)
thì mình bảo câu 1 làm tươgn tự nhé, còn câu 3 ko làm tương tự đc, hiểu ko bạn Thanh Hằng Nguyễn
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) \(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\\dfrac{3}{4}-1\dfrac{1}{2}x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\\dfrac{3}{2}x=\dfrac{3}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}.3=-\dfrac{3}{2}\\x=\dfrac{3}{4}:\dfrac{3}{2}=\dfrac{1}{2}\end{matrix}\right.\)