Tìm x biết \(\left|5\left(2x+3\right)\right|+\left|2\left(2x+3\right)\right|+\left|2x+3\right|=16\)
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1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
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`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
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`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
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`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
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`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
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`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
Ta có: \(\left|5\left(2x+3\right)\right|+\left|2\left(2x+3\right)\right|+\left|2x+ 3\right|=16\)
\(\Rightarrow5\left|2x+3\right|+2\left|2x+3\right|+\left|2x+3\right|=16\)
\(\Rightarrow\left|2x+3\right|\left(5+2+1\right)=16\)
\(\Rightarrow\left|2x+3\right|.8=16\)
\(\Rightarrow\left|2x+3\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=2\\2x+3=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-1\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\).
a) Ta có \(|5\left(2x+3\right)\ge0\)
\(|2\left(2x+3\right)|\ge0\)
\(|2x+3|\ge0\)
\(\Rightarrow|5\left(2x+3\right)|+|\left(2x+3\right)|+|2x+3|\ge0\)
\(\Rightarrow5\left(2x+3\right)+2\left(2x+3\right)+2x+3=16\)
\(\Rightarrow10x+15+4x+6+2x+3=16\)
\(\Rightarrow\left(10x+4x+2x\right)+\left(15+6+3\right)=16\)
\(\Rightarrow16x+24=16\)
\(\Rightarrow24=16x-16\)
\(\Rightarrow24=x\)
Vậy x=24
a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-\left(x^2-x+x-1\right)=16\)
\(\Leftrightarrow8x+1=0\Leftrightarrow x=-\frac{1}{8}\)
b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)
c, \(\left(x+2\right)\left(x-2\right)-x^3-2x=15\)
\(\Leftrightarrow x^2-4-x^3-2x=15\)( vô nghiệm )
d, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1=28\)
\(\Leftrightarrow15x^2+26=0\Leftrightarrow x^2\ne-\frac{26}{15}\)( vô nghiệm )
Tính nhẩm hết á, sai bỏ quá nhá, sắp đi hc ... nên chất lượng hơi kém xíu ~~~
NX: 2x+3; 5(2x+3) và 2(2x+3) cùng dấu
+TH1: 2x+3 \(\ge\)0 => x \(\ge\frac{-3}{2}\)
=> 5(2x+3), 2(2x+3) \(\ge\)0
=> |5(2x+3)| = 5(2x+3); |2(2x+3)| = 2(2x+3); |2x+3| = 2x+3
=> (2x+3)(5+2+1) = 16
=> 2x+3 = 2
=> 2x = -1
=> x = -1/2 (t/m)
+ TH2: 2x+3 < 0 => x < -3/2
cmtt => -5(2x+3) - 2(2x+3) - (2x+3) = 16
=> (2x+3)(-5-2-1) = 16
=> 2x+3 = -2
=> 2x = -5
=> x = -5/2 (t/m)
/8(2x+3/ = 16
/2x+3/=2
2x+3=2 hoặc 2x+3=-2
2x=-1 hoặc 2x=-5
x=-1/2 hoặc x=-5/2
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