Tìm x biết:
a) 7x+2 + 2.7x-1= 345
b) 2x+2 - 2x= 96
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e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
a,(2x-5^2)-4x(x-3)=0
=> 2x-25-4x2+12x=0
=>-4x2+14x-25=0
đề bài ý a sai nha
b, 6x2-7x=0
=>x(6x-7)=0
=>x=0 và 6x-7=0
=>x=0 và x=7/6
vậy x=0 và x=7/6
`a)|2x-15|=13`
`**2x-15=13`
`<=>2x=28`
`<=>x=14.`
`**2x-15=-13`
`<=>2x=-2`
`<=>x=-1.`
`b)|7x+3|=66`
`**7x+3=66`
`<=>7x=63`
`<=>x9`
`**7x+3=-66`
`<=>7x=-69`
`<=>x=-69/7`
`c)|5x-2|=0`
`<=>5x-2=0`
`<=>5x=2`
`<=>x=2/5`
\(a,\Leftrightarrow\left[{}\begin{matrix}2x-5=13\\2x-5=-13\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)
Vậy ...
\(b,\Leftrightarrow\left[{}\begin{matrix}7x+3=66\\7x+3=-66\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{69}{7}\end{matrix}\right.\)
Vậy ...
\(c,\Leftrightarrow5x-2=0\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy ...
a: Ta có: \(7x+25=144\)
\(\Leftrightarrow7x=119\)
hay x=17
b: Ta có: \(33-12x=9\)
\(\Leftrightarrow12x=24\)
hay x=2
c: Ta có: \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
hay x=31
d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)
\(\Leftrightarrow5\left(726-3x\right)=2175\)
\(\Leftrightarrow726-3x=435\)
\(\Leftrightarrow3x=291\)
hay x=97
e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x+5\right)=18\)
\(\Leftrightarrow2x+5=23\)
\(\Leftrightarrow2x=18\)
hay x=9
a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)
\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)
\(\Leftrightarrow-12x=12\)
hay x=-1
Bài 2: Tìm số tự nhiên x biết:
a) 12(x−1)∶3=43+23 c) (7x−11)3=23.52+200
b) 128−3(x+4)=23 d) 2x+2−2x=96
a) PT \(\Leftrightarrow x^2-x-x^2+2x=5\) \(\Rightarrow x=5\)
Vậy ...
b) PT \(\Leftrightarrow8x=16\) \(\Rightarrow x=2\)
Vậy ...
a: Ta có: \(x\left(x-1\right)-x^2+2x=5\)
\(\Leftrightarrow x^2-x-x^2+2x=5\)
hay x=5
b: Ta có: \(2x\left(3x+4\right)-6x^2=16\)
\(\Leftrightarrow6x^2+8x-6x^2=16\)
\(\Leftrightarrow8x=16\)
hay x=2
a: ĐKXĐ: x<>-1/2
\(\dfrac{x-1}{2x+1}=\dfrac{2}{3}\)
=>\(2\left(2x+1\right)=3\left(x-1\right)\)
=>\(4x+2=3x-3\)
=>\(4x-3x=-3-2\)
=>x=-5(nhận)
b: ĐKXĐ: x<>1/2
\(\dfrac{x-2}{2x-1}=\dfrac{-1}{3}\)
=>\(3\left(x-2\right)=-1\left(2x-1\right)\)
=>\(3x-6=-2x+1\)
=>\(3x+2x=1+6\)
=>5x=7
=>x=7/5(nhận)
a.
\(7^{x+2}+2\times7^{x-1}=345\)
\(7^x\times7^2+2\times7^x\div7=345\)
\(7^x\times\left(7^2+\frac{2}{7}\right)=345\)
\(7^x\times\frac{345}{7}=345\)
\(7^x=345\div\frac{345}{7}\)
\(7^x=345\times\frac{7}{345}\)
\(7^x=7\)
\(x=1\)
b.
\(2^{x+2}-2^x=96\)
\(2^x\times\left(2^2-1\right)=96\)
\(2^x\times3=96\)
\(2^x=\frac{96}{3}\)
\(2^x=32\)
\(2^x=2^5\)
\(x=5\)
\(a,7^{x+2}+2.7^{x-1}=345\Rightarrow7^x.49+\frac{2}{7}.7^x=345\Rightarrow7^x\left(49+\frac{2}{7}\right)=345\Rightarrow7^x.\frac{345}{7}=345\Rightarrow7^x=345:\frac{345}{7}=7^1\Rightarrow x=1\)
\(b,2^{x+2}-2^x=96\Rightarrow2^x.4-2^x=96\Rightarrow2^x\left(4-1\right)=96\Rightarrow2^x.3=96\Rightarrow2^x=96:3=32\Rightarrow2^x=2^5\Rightarrow x=5\)