tìm x
a,[x-1954] *x=50
b,3[*[x+2]/7]*4
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a: x chia hết cho 4;5;10
nên \(x\in BC\left(4;5;10\right)\)
mà 10<=x<50
nên x=40
b: x=33
a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)
\(\Leftrightarrow-62x=-92\)
hay \(x=\dfrac{46}{31}\)
b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)
\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)
\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)
\(\Leftrightarrow8x^2+49x-15=0\)
\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)
`@` `\text {Ans}`
`\downarrow`
`a,`
`(x-1954) \times 5=50`
`x-1954 = 50 \div 5`
`x-1954 = 10`
`x=10 + 1954`
`x= 1964`
`B.`
`48 - 40 \times x=8`
`40 \times x=48-8`
`40 \times x = 40`
` x=40 \div 40`
`x=1`
`C.`
`2+70 \div x = 3`
`70 \div x = 3-2`
`70 \div x = 1`
`x=70 \div 1`
`x=70`
`D. [3 \times (x+2) \div 7] \times 4=120`
`3 \times (x+2) \div 7 = 120 \div 4`
`3 \times (x+2) \div 7 =30`
`3 \times (x+2) = 30 \times 7`
`3 \times (x+2)=210`
`x+2=210 \div 3`
`x+2=70`
`x=70-2`
`x=68`
`E.`
`7,2 \div [ (0,6 \times x+8) \div 20 + 59] = 0,12?` Thiếu ngoặc bạn nhé._.
\(a,\left(x-1954\right)\times5=50\\ x-1954=50:5\\ x-1954=10\\ x=1954+10\\ x=1964\\b.48-40\times x=8\\40\times x=48-8\\40\times x=40\\ x=40:40\\ x=1 \\ c.2+70:X=3\\ 70:x=3-2\\ 70:x=1\\ x=70:1\\ x=70\\ d,\left[3\times\left(x+2\right):7\right]\times4=120\\ 3\times\left(x+2\right):7-120:4\\ 3\times\left(x+2\right):7=30\\ \left(x+2\right):7=30:3\\ \left(x+2\right):7=10\\ x+2=10\times7\\ x+2=70\\ x=70-2\\ x=68\)
a, Ta có :
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)
\(\Rightarrow x=11;y=17;z=23\)
b, Đặt \(\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\Rightarrow xyz=810\)
\(\Rightarrow2k.3k.5k=810\Leftrightarrow30k^3=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)
\(\Rightarrow x=6;y=9;z=15\)
a) Ta có: \(\dfrac{x-1}{2}=\dfrac{2x-2}{4};\dfrac{y-2}{3}=\dfrac{3y-6}{9};\dfrac{z-3}{4}\)
Áp dụng t/c dtsbn:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=5\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=12\end{matrix}\right.\)
b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
xyz = 810
=> 2k.3k.5k = 810
=> k = 3
\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)
Bài 5:
a)x+37=50
⇔x=13
b)2x-3=11
⇔2x=14
⇔x=7
c)(2+x):5=6
⇔2+x=30
⇔x=28
d)2+x:5=6
⇔x:5=4
⇔x=20
a)x+37=50
x =50-37
x =13
b)2.x-3=11
2.x =11+3
2.x =14
x =14:2
x =7
c)(2+x):5=6
2+x =6.5
2+x =30
x=30-2
x=28
d)2+x:5=6
x:5=6-2
x:5=4
x=4.5
x=20
Câu trả lời nhớ ghi đầy đủ như này nha bạn
A) \(x-\dfrac{2}{3}=\dfrac{4}{5}\\ x=\dfrac{4}{5}+\dfrac{2}{3}\)
\(x=\dfrac{22}{15}\)
b)\(\dfrac{7}{9}-x=\dfrac{1}{3}\\ x=\dfrac{7}{9}-\dfrac{1}{3}\\ x=\dfrac{4}{9}\)
C)\(x:\dfrac{2}{3}=\dfrac{9}{8}\\ x=\dfrac{9}{8}x\dfrac{2}{3}\\ x=\dfrac{3}{4}\)
a: x=2/5*15=6
b: =>x-7=15
=>x=22
c: =>3/4:x+1/2*4=4
=>3/4:x=4-2=2
=>x=3/8
a,3/4 . (-5/12)+3/4.(-7/12)
` 3/4 . [ - ( 5/12 + 7/12 ) ] `
`3/4 . (-1) = -3/4 `
`2/3 . x - 0,5 = 3/4 `
` x - 0,5 = 3/4 - 2/3 `
` x-0,5 = 1/12 `
` x = 1/12 + 0,5 `
` x= 7/12 `