2/5 x 1/x + 1/x X 2 = 0,1
tìm x
câu 1 . 150 + x : 3=620 :4
câu 2 X x 0,1 = 1/2 - 2/5
câu 3 4/9 +5/9 : x =1
lưu ý câu 2 là ích x 0,1 ....... nhé
a.
150 + x : 3 = 620 : 4
150 + x : 3 = 155
x : 3 = 155 - 150
x : 3 = 5
x = 5 x 3
x = 15
b.
x * 0,1 = 1/2 -2/5
x * 1/10 = 5/10 - 4/10
x * 1/10 = 1/10
x = 1/10 : 1/10
x = 1
c.
4/9 + 5/9 : x = 1
5/9 : x = 1 - 4/9
5/9 :x = 5/9
x = 5/9 : 5/9
x = 1
1) \(150+x.\frac{1}{3}=155\)
\(\frac{1}{3}x=5\)
\(x=15\)
2) \(x.\frac{1}{10}=\frac{1}{2}-\frac{2}{5}\)
\(\frac{1}{10}x=\frac{5}{10}-\frac{4}{10}\)
\(\frac{1}{10}x=\frac{1}{10}\)
\(x=1\)
3) \(\frac{4}{9}+\frac{9}{5}x=1\)
\(\frac{9}{5}x=\frac{9}{9}-\frac{4}{9}\)
\(\frac{9}{5}x=\frac{5}{9}\)
\(x=1\)
1) 2x+3x+5x. 2) 2.x- x + 3.x. 3) 9.x - 3- 3.x
4) 0,8. x 0,2x-0,1.x. 5) x-0,2x -0,1.x. 6) 7/2 x-1/2 x
Giúp em với nhanh nhất vì tối nay em đi học rồi ạ
Tính phải k nhỉ?
`1)`
`2x + 3x + 5x`
`= (2 + 3 + 5)x`
`= 10x`
`2)`
`2.x - x + 3.x`
`= (2 - 1 + 3)x`
`= 4x`
`3)`
`9.x - 3 - 3.x`
`= (9 - 3)x - 3`
`= 6x - 3`
`4)`
Thiếu dấu, bạn bổ sung thêm
`5)`
`x - 0,2x - 0,1x`
`= (1 - 0,2 - 0,1)x`
`=0,7x`
`6)`
\(\dfrac{7}{2}x-\dfrac{1}{2}x=\left(\dfrac{7}{2}-\dfrac{1}{2}\right)x=3x\)
Giải bất phương trình:
\(a,\log_{0,1},1\left(x^2+x-2\right)>\log_{0,1}\left(x+3\right)\)
\(b,\log_{\dfrac{1}{3}}\left(x^2-6x+5\right)+2\log_3\left(2-x\right)\ge0\)
a. Vì \(0< 0,1< 1\) nên bất phương trình đã cho
\(\Leftrightarrow0< x^2+x-2< x+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x-2>0\\x^2-5< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< -2\\x>1\end{matrix}\right.\\-\sqrt{5}< x< \sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{5}< x< -2\\1< x< \sqrt{5}\end{matrix}\right.\)
Vậy tập nghiệm của bất phương trình là \(S=\left\{-\sqrt{5};-2\right\}\) và \(\left\{1;\sqrt{5}\right\}\)
b. Điều kiện \(\left\{{}\begin{matrix}2-x>0\\x^2-6x+5>0\end{matrix}\right.\)
Ta có:
\(log_{\dfrac{1}{3}}\left(x^2-6x+5\right)+2log^3\left(2-x\right)\ge0\)
\(\Leftrightarrow log_{\dfrac{1}{3}}\left(x^2-6x+5\right)\ge log_{\dfrac{1}{3}}\left(2-x\right)^2\)
\(\Leftrightarrow x^2-6x+5\le\left(2-x\right)^2\)
\(\Leftrightarrow2x-1\ge0\)
Bất phương trình tương đương với:
\(\left\{{}\begin{matrix}x^2-6x+5>0\\2-x>0\\2x-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>5\end{matrix}\right.\\x< 2\\x\ge\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{2}\le x< 1\)
Vậy tập nghiệm của bất phương trình là: \(\left(\dfrac{1}{2};1\right)\)
a,x.(4/5.x-1),(0,1.x-10)=0 b,(1/4.x-1)-(5/6.x+2)-(1-5/8.x)=0
Lời giải:
a. $x(\frac{4}{5}x-1)(0,1x-10)=0$
$\Rightarrow x=0$ hoặc $\frac{4}{5}x-1=0$ hoặc $0,1x-10=0$
Nếu $\frac{4}{5}x-1=0$
$\Rightarrow x=1: \frac{4}{5}=\frac{5}{4}$
Nếu $0,1x-10=0$
$\Rightarrow x=10:0,1=100$
Vậy $x=0; \frac{5}{4}; 100$
b.
$(\frac{1}{4}x-1)-(\frac{5}{6}x+2)-(1-\frac{5}{8}x)=0$
$(\frac{1}{4}x-\frac{5}{6}x+\frac{5}{8}x)-(1+2+1)=0$
$\frac{1}{24}x-4=0$
$x=4: \frac{1}{24}=96$
X : 0,1 +
\(\Leftrightarrow x\cdot\dfrac{21}{2}=\dfrac{3}{7}\)
hay x=2/49
Giải phương trình sau :
a) 11 + 8x – 3 = 5x – 3 + x
b) 2x(x + 2)² - 8x² = 2(x – 2)(x² + 2x + 4)
c) (x + 1)(2x – 3) = (2x – 1)(x + 5)
d) 0,1 – 2(0,5t – 0,1) = 2(t – 2,5) – 0,7
a: Ta có: \(8x+11-3=5x+x-3\)
\(\Leftrightarrow8x+8=6x-3\)
\(\Leftrightarrow2x=-11\)
hay \(x=-\dfrac{11}{2}\)
b: Ta có: \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow2x\left(x^3+6x^2+12x+8\right)-8x^2=2\left(x^3-8\right)\)
\(\Leftrightarrow2x^4+12x^3+24x^2+16x-8x^2-2x^3+16=0\)
\(\Leftrightarrow2x^4+10x^3+16x^2+16x+16=0\)
\(\Leftrightarrow2x^4+4x^3+6x^3+12x^2+4x^2+8x+8x+16=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x^3+6x^2+4x+8\right)=0\)
\(\Leftrightarrow x+2=0\)
hay x=-2
c: Ta có: \(\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\)
\(\Leftrightarrow2x^2-3x+2x-3-2x^2-10x+x+5=0\)
\(\Leftrightarrow-10x+2=0\)
\(\Leftrightarrow-10x=-2\)
hay \(x=\dfrac{1}{5}\)
d: Ta có: \(\dfrac{1}{10}-2\cdot\left(\dfrac{1}{2}t-\dfrac{1}{10}\right)=2\left(t-\dfrac{5}{2}\right)-\dfrac{7}{10}\)
\(\Leftrightarrow\dfrac{1}{10}-t+\dfrac{1}{5}=2t-5-\dfrac{7}{10}\)
\(\Leftrightarrow-t-2t=-\dfrac{57}{10}-\dfrac{3}{10}=-6\)
hay t=2
Tìm \(x\) biết:
\(\left(\sqrt{3}\right)^x=243\)
\(0,1^x=1000\)
\(\left(\dfrac{1}{2}\right)^x=1024\)
\(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)
\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)
\(5^{x-1}+5^{x+2}=3\)
a: \(\left(\sqrt{3}\right)^x=243\)
=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)
=>\(\dfrac{1}{2}\cdot x=5\)
=>x=10
b: \(0,1^x=1000\)
=>\(\left(\dfrac{1}{10}\right)^x=1000\)
=>\(10^{-x}=10^3\)
=>-x=3
=>x=-3
c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)
=>\(\left(0,2\right)^{x+3}< 0,2\)
=>x+3>1
=>x>-2
d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)
=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)
=>2x+1<-2
=>2x<-3
=>\(x< -\dfrac{3}{2}\)
e: \(5^{x-1}+5^{x+2}=3\)
=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)
=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)
=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)
a) 1/5+1/2+3/10<X<3+1/3+1/2+1/6
b)1000x0,1<X<10,3:0,1
\(a,\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{3}{10}=\dfrac{2+5+3}{10}=\dfrac{10}{10}=1\\ 3+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{3\times6+2+3+1}{6}=4\\ Vậy:1< x< 4\\ Vậy:x=2.hoặc.x=3\)
\(b,1000\times0,1=100\\ 10,3:0,1=103\\ Vậy:100< x< 103\\ Vậy:x=101.hoặc.x=102\)
Giải các phương trình sau
A.5+3x=4x-9
B.3,2x-5(x-0,2)=5+0,2x
C.1,5-(x+2)=-3(x+0,1)
E.2/3-1/2(x+2)=-x+1
F.3t-4+13+2(t+2)-3t
A 3x-4x=-9-3
-x=-12
x=12
B 3.2x -5x +1=5+0.2x
3.2x-5x-0.2x=5-1
-2x=4
x=-2
C 1.5-x-2=-3x-0.3
-x+3x=-0.3-1.5+2
2x =0.2
x=0.1
E 2/3-1/2x-1=-x+1
-1/2x+x=1+1-2/3
1/2x=4/3
x=8/3
F 3t-4+13+2t+4-3t
=3t+2t-3t-4+13+4
=2t+13