Tìm x biết \(2\times x-3=5\)
TÌM X BIẾT: \(\frac{1}{2}\times x+\frac{3}{5}\times\left(x-2\right)=3\)
1/2x + 3/5(x-2) = 3
\(\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)
\(\frac{11}{10}x=\frac{21}{5}\)
\(x=\frac{21}{5}:\frac{11}{10}\)
\(x=\frac{42}{11}\)
#mã mã#
1/2x + 3/5x - 6/5 = 3
x(1/2 + 3/5) = 3 + 6/5
11/10x = 21/5
x = 21/5 : 11/10
x = 42/11
\(\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)
\(\Rightarrow\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{3}{5}\right)-\frac{6}{5}=3\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{3}{5}\right)=\frac{21}{5}\)
\(\Rightarrow\frac{11}{10}x=\frac{21}{5}\Rightarrow x=\frac{21}{5}\div\frac{11}{10}=\frac{42}{11}\)
Vậy \(x=\frac{42}{11}\)
Câu 1: Tìm x biết
1,\(5^2\times7^3\times11^2\times x-5^2\times7^2\times11^4=0\)
2,\(5^2\times7^3\times11^2\times x+5^3\times7^2\times11=0\)
vì số nào nhân vs o đều bàng 0 nên x trong cả hai bài trên đều bằng 0
1, x=0
2, x=0
tìm x biết
\(x\times\left(4.6+\frac{3}{5}\right)+7.2-8.15\)\(42+\frac{3}{7}\times\left[3\times-1\right]=12\) \(\left[2\times x-1\right]=\left[x+2\right]\) \(\frac{x+2}{5}=\frac{2-3\times x}{3}\)a) Thiếu đề (hoặc sai)
b) x đâu?
c)\(3x-1=x+2\)
\(\Rightarrow3x-x=2+1\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\frac{3}{2}\)
c) \(\frac{x+2}{5}=\frac{2-3x}{3}\)
\(\Rightarrow3.\left(x+2\right)=5.\left(2-3x\right)\)
\(\Rightarrow3x+6=10-15x\)
\(\Rightarrow3x+15x=10-6\)
\(\Rightarrow18x=4\)
\(\Rightarrow x=\frac{4}{18}=\frac{2}{9}\)
câu 1 là \(x\times\left(4.6+\frac{3}{5}\right)=7.2-8.15\)
câu 2 là \(42+\frac{3}{7}.\left[3\times x-1=12\right]\)
a)\(x.\left(4,6+\frac{3}{5}\right)=7,2-8,15\)
\(\Rightarrow\left(\frac{23}{5}+\frac{3}{5}\right).x=-0,95\)
\(\Rightarrow\frac{26}{5}.x=\frac{-19}{20}\)
\(\Rightarrow x=\frac{-19}{20}:\frac{26}{5}\)
\(\Rightarrow x=\frac{-19}{104}\)
Mà \(4.6\)là\(4,6\)hay \(4\times6?\)nếu sai tớ làm lại
b) Ngoặc sai rồi
\(42+\frac{3}{7}.\left(3x-1\right)=12\)
\(\Rightarrow\frac{3}{7}.\left(3x-1\right)=12-42\)
\(\Rightarrow\frac{3}{7}.\left(3x-1\right)=-30\)
\(\Rightarrow3x-1=-30:\frac{3}{7}\)
\(\Rightarrow3x-1=\frac{-10}{7}\)
\(\Rightarrow3x=\frac{-10}{7}+1\)
\(\Rightarrow3x=\frac{-10+7}{7}=\frac{-3}{7}\)
\(\Rightarrow x=\frac{-3}{7}:3=\frac{-1}{7}\)
Lưu ý nhé viết dấu nhân thì viết \("."\)một thể, \("\times"\)một thể. Còn dâu phảy của số thập phân thì viết ",'' nhé
tìm x \(\in\) Q biết rằng
\(\dfrac{11}{12}\) - ( \(\dfrac{2}{5}\) + x ) = \(\dfrac{2}{3}\)
2x \(\times\) ( x - \(\dfrac{1}{7}\) ) = 0
\(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) : x = \(\dfrac{2}{5}\)
1) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{3}{20}\)
2) \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
3) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4x}=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4x}=-\dfrac{7}{20}\)
\(\Leftrightarrow4x=-\dfrac{20}{7}\)
\(\Leftrightarrow x=-\dfrac{5}{7}\)
tìm x , biết :
\(\frac{4}{5}x^2\times\left(\frac{x}{3}-\frac{1}{2}\right)-\left(\frac{1}{5}x-\frac{2}{3}\right)\times\left(\frac{4x^2}{3}+1\right)=\frac{22}{45}x^2\)
\(\Leftrightarrow\frac{4x^2}{5}\times\frac{2x-3}{6}-\frac{3x-10}{15}\times\frac{4x^2+3}{3}=\frac{22x^2}{45}\)
\(\Leftrightarrow\frac{4x^2\left(2x-3\right)}{30}-\frac{\left(3x-10\right)\left(4x^2+3\right)}{45}=\frac{22x^2}{45}\)
\(\Leftrightarrow\frac{12x^2\left(2x-3\right)}{90}-\frac{2\left(3x-10\right)\left(4x^2+3\right)}{90}=\frac{44x^2}{90}\)
\(\Leftrightarrow12x^2\left(2x-3\right)-2\left(3x-10\right)\left(4x^2+3\right)=44x^2\)
\(\Leftrightarrow24x^2-36x^2-2\left(12x^3+9x-40x^2-30\right)=44x^2\)
\(\Leftrightarrow24x^2-36x^2-24x^3-18x+80x^2+60=44x^2\)
\(\Leftrightarrow24x^3-36x^2-24x^3-18x+80x^2-44x^2=-60\)
\(\Leftrightarrow\left(24x^3-24x^3\right)+\left(-36x^2+80x^2-44x^2\right)-18x=-60\)
\(\Leftrightarrow-18x=-60\)
\(\Leftrightarrow x=\frac{-60}{-18}\)
\(\Leftrightarrow x=\frac{10}{3}\)
Tìm x biết
a, | 1/2 × x - 3 | = 1/2
b, | 3× x + 1/5 | - 1/3 = 2/3
c, |3× x - 1/5 | + | 15× x - 1 |= 6/2
d, | 1/2 × x - 3 | < 1/2
e, | 3× x - 4/5 | > 1/3
a: \(\left|\dfrac{1}{2}x-3\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=\dfrac{1}{2}\\\dfrac{1}{2}x-3=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{7}{2}\\\dfrac{1}{2}x=\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=5\end{matrix}\right.\)
b: \(\left|3x+\dfrac{1}{5}\right|-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\Leftrightarrow\left|3x+\dfrac{1}{5}\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{1}{5}=1\\3x+\dfrac{1}{5}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{4}{5}\\3x=-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{15}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d: \(\left|\dfrac{1}{2}x-3\right|< \dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-3>-\dfrac{1}{2}\\\dfrac{1}{2}x-3< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x>\dfrac{5}{2}\\\dfrac{1}{2}x< \dfrac{7}{2}\end{matrix}\right.\Leftrightarrow5< x< 7\)
e: \(\left|3x-\dfrac{4}{5}\right|>\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{4}{5}>\dfrac{1}{3}\\3x-\dfrac{4}{5}< -\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x>\dfrac{17}{15}\\3x< \dfrac{7}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{17}{45}\\x< \dfrac{7}{45}\end{matrix}\right.\)
Tìm x , biết
a, -12 × ( x - 5 ) + 7 × ( 3 × x ) = 5
b, 30 × ( x + 2 ) - 6 × ( x - 5 ) - 24 × x = 100
c, | 5 × x - 2 | < 13 ( x thuộc Z )
a) -12.(x-5)+7.(3.x)=5
<=> -12x+60+21+7x=5
<=>-5x+81=5
<=>-5x=5-81=-76
<=>x=-76/-5=76/5=15,2
b) 30.(x+2)-6.(x-5)-24.x=100
<=> 30x+60-6x+30-24x=100
<=> 0x=100-60-30=10
=> không có giá trị nào của x để 0x=10
c) \(|5.x-2|< 13\)
Khi 5x-2 < 13
<=> 5x<15 <=> x<3
Khi 5x-2 <-13
<=> 5x<-11 <=> x<-11/5 <=> x<-2,2
Tìm x;y;z biết:
\(x:y:z=3:4:5\) và \(2\times x^2+2\times y^2-3\times z^2=-100\)
Giúp mình với mình cần gấp!
Ta có: x/3=y/4=z/5.......
2*x^2/2*3^2+2*y^2/2*4^2-3*z^2=-100/-25=4
x/3=4 suy ra x=12
y/4=4 ....y=16
z/5.......z=20
Ta co : x:y:z=3:4:5
Hay : x/3=y/4=z/5
=>2x^2/18=2y^2/32=3z^2/75 và 2x^2+2y^2-3z^2=-100
Áp dụng tính chất dãy tỉ số bằng nhau :
2x^2/18=2y^2/32=3z^2/75=2x^2+2y^2-3z^2/18+32-75=-100/-25=4
Suy ra : 2x^2/18=4=>2x^2=72=>x^2=36=>x=+6
2y^2/32=4=>2y^2=128=>y^2=64=>y=+8
3z^2/75=4=>3z^2=300=>z^2=100=>z=+10
k nha , k hiu ns mk
Tìm XX biết: (X \times 5 - 19) + (X \times 3 + 24) = 125.(X×5−19)+(X×3+24)=125. \Trả lời: X =X=