( x + 2 ) . 16 . x = 160 . x
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a;80x-64=160
80x=224
x=14/5
b;x-4=42:6
x-4=7
x=11
c;40-x=16*2
40-x=32
x=8
d;20-x=8:4
20-x=2
x=18
`# \text {Ryo}`
`a)`
`80x - 2^2 . 2^4 = 160`
`\Rightarrow 80x - 2^6 = 160`
`\Rightarrow 80x - 64 = 160`
`\Rightarrow 80x = 224`
`\Rightarrow x = 224 \div 80`
`\Rightarrow x = 2,8`
Vậy, `x = 2,8`
`b)`
`(x - 4).6 = 42`
`\Rightarrow x - 4 = 42 \div 6`
`\Rightarrow x - 4 = 7`
`\Rightarrow x = 7 + 4`
`\Rightarrow x = 11`
Vậy, `x = 11`
`c)`
`(40 - x) \div 2 = 16`
`\Rightarrow 40 - x = 16 . 2`
`\Rightarrow 40 - x = 32`
`\Rightarrow x = 40 - 32`
`\Rightarrow x = 8`
Vậy, `x = 8`
`d)`
`(20 - x) . 4 = 8`
`\Rightarrow 20 - x = 8 \div 4`
`\Rightarrow 20 - x = 2`
`\Rightarrow x = 20 - 2`
`\Rightarrow x = 18`
Vậy, `x = 18.`
\(\left(84,6-2\cdot x\right):3,02=5,1\)
\(\Rightarrow84,6-2\cdot x=15,402\)
\(\Rightarrow2\cdot x=69,198\)
\(\Rightarrow x=69,198:2\)
\(\Rightarrow x=34,599\)
_____________
\(\left(15\cdot24-x\right):0,25=100:0,25\)
\(\Rightarrow\left(360-x\right):0,25=400\)
\(\Rightarrow360-x=100\)
\(\Rightarrow x-360-100\)
\(\Rightarrow x=260\)
______________
\(128\cdot x-12\cdot x-16\cdot x=5200\)
\(\Rightarrow x\cdot\left(128-12-16\right)=5200\)
\(\Rightarrow x\cdot100=5200\)
\(\Rightarrow x=5200:100\)
\(\Rightarrow x=52\)
__________________
\(5\cdot x+3,75\cdot x+1,25\cdot x=20\)
\(\Rightarrow x\cdot\left(5+3,75+1,25\right)=20\)
\(\Rightarrow10\cdot x=20\)
\(\Rightarrow x=20:10\)
\(\Rightarrow x=2\)
\(x\cdot3,7+x\cdot6,3=360:120\)
\(\Rightarrow x\cdot\left(3,7+6,3\right)=3\)
\(\Rightarrow x\cdot10=3\)
\(\Rightarrow x=\dfrac{3}{10}\)
__________________
\(x\cdot23-6\cdot23+x\cdot69=320\)
\(\Rightarrow x\cdot\left(23+69\right)=320+6\cdot23\)
\(\Rightarrow x\cdot92=458\)
\(\Rightarrow x=458:92\)
\(\Rightarrow x=\dfrac{229}{46}\)
___________________
\(\left(x+1\right)\left(x+2\right)=72\)
\(\Rightarrow x^2+2x+x+2=72\)
\(\Rightarrow x^2+3x+2=72\)
\(\Rightarrow x^2+3x+2-72=0\)
\(\Rightarrow x^2+3x-70=0\)
\(\Rightarrow x^2+10x-7x-70=0\)
\(\Rightarrow\left(x-7\right)\left(x+10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-10\end{matrix}\right.\)
___________________
\(\left(x+2\right)\cdot16\cdot x=160x\)
\(\Rightarrow16x^2+32x=160x\)
\(\Rightarrow16x^2+32x-160x=0\)
\(\Rightarrow16x^2-128x=0\)
\(\Rightarrow16x\left(x-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
\(\left(x+16\right)\left(x+56\right)=160\)
\(\Leftrightarrow x^2+56x+16x+896=160\)
\(\Leftrightarrow x^2+72x+896=160\)
\(\Leftrightarrow x^2+72x+1296-400=160\)
\(\Leftrightarrow\left(x+36\right)^2=560\)
\(\Leftrightarrow\orbr{\begin{cases}x+36=\sqrt{560}\\x+36=-\sqrt{560}\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\sqrt{35}-36\\x=-4\sqrt{35}-36\end{cases}}\)
a) \(6x+15\times8=12\times\left(19-x\right)\)
\(6x+120=228-12x\)
\(6x+120-228+12x=0\)
\(18x-108=0\)
\(18x=108\)
\(x=6\)
b) \(160-\left(35\div x+3\right)\times15=15\)
\(160-\left(35\div x+3\right)=1\)
\(35\div x+3=159\)
\(35\div x=156\)
\(x=\dfrac{35}{156}\)
c) \(2x-\left(1309\div11-19\right)-2=0\)
\(2x-1309\div11-19=2\)
\(2x-119-19=2\)
\(2x-119=21\)
\(2x=140\)
\(x=70\)
d) \(\left(x-7\right)\times\left(2x-16\right)=0\)
\(x-7=0;2x-16=0\)
\(x=7;2x=16\)
\(x=7;x=8\)
a) \(\left(-5\right)\left(x-2\right)^2+360=\left(-150\right)\cdot3+43\cdot5\)
\(-5\cdot\left(x-2\right)^2+360=-235\)
\(-5\cdot\left(x-2\right)^2=-595\)
\(\left(x-2\right)^2=119\)
\(\left(x-2\right)^2=\left(\pm\sqrt{199}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-2=\sqrt{119}\\x-2=-\sqrt{119}\end{cases}\Rightarrow\orbr{\begin{cases}x=\sqrt{119}+2\\x=2-\sqrt{119}\end{cases}}}\)
b) \(\left(x+5\right)-\left(3x+9\right)=-16\)
\(x+5-3x-9=-16\)
\(-2x-4=-16\)
\(-2x=-12\)
\(x=6\)
c) \(3\left(x+2\right)-\left(15-x\right)\cdot6=160+\left(-1\right)^{1005}\)
\(3x+6-90+6x=160-1\)
\(9x-84=159\)
\(9x=243\)
\(x=27\)
d) \(x\left(x-1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x\left(x-1\right)=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\left\{0;1\right\}\\x=\left\{2;-2\right\}\end{cases}}}\)
\(PTK_{CuSO_x}=NTK_{Cu}+NTK_S+x\cdot NTK_O=160\\ \Rightarrow64+32+16x=160\\ \Rightarrow16x=64\\ \Rightarrow x=4\\ \Rightarrow A\)
a) ADTCDTSBN
có: \(\frac{x}{12}=\frac{y}{13}=\frac{z}{15}=\frac{x+y+z}{12+13+15}=\frac{160}{40}=4\)
=> x/12 = 4 => x = 48
...
b) ta có: \(x=\frac{y}{6}=\frac{z}{3}=\frac{2x}{2}=\frac{3y}{18}=\frac{4z}{12}\)
ADTCDTSBN
có: \(\frac{2x}{2}=\frac{3y}{18}=\frac{4z}{12}=\frac{2x-3y+4z}{2-18+12}=\frac{16}{-4}=-4\)
=>...
c) ta có: \(\frac{x}{2}=\frac{y}{-3}=\frac{z}{3}=\frac{2x}{4}=\frac{3y}{-9}=\frac{2z}{8}\)
ADTCTDBN
có: \(\frac{2x}{4}=\frac{3y}{-9}=\frac{2z}{8}=\frac{2x+3y+2z}{4-9+8}=\frac{1}{3}\)
=>...
( x + 2 ) . 16 . x = 160 . x
( x + 2 ) . 16 . x = 16 . 10 . x
x + 2 = 10
x = 8
( x + 2 ) . 16 . x = 160 . x
( x + 2 ) . 16 . x = 16 . 10 . x
x + 2 = 10
x = 8