Tìm x biết
a ( 2x - 1 ) 3 - 25 = 100
b 2*3 x-1 - 15 =3
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\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)
Bài 1.
1) ( 2x + 1 )3 - ( 2x + 1 )( 4x2 - 2x + 1 ) - 3( 2x - 1 ) = 15
<=> 8x3 + 12x2 + 6x + 1 - [ ( 2x )3 - 13 ] - 6x + 3 = 15
<=> 8x3 + 12x2 + 4 - 8x3 + 1 = 15
<=> 12x2 + 15 = 15
<=> 12x2 = 0
<=> x = 0
2) x( x - 4 )( x + 4 ) - ( x - 5 )( x2 + 5x + 25 ) = 13
<=> x( x2 - 16 ) - ( x3 - 53 ) = 13
<=> x3 - 16x - x3 + 125 = 13
<=> 125 - 16x = 13
<=> 16x = 112
<=> x = 7
Bài 2.
A = ( x + 5 )( x2 - 5x + 25 ) - ( 2x + 1 )3 - 28x3 + 3x( -11x + 5 )
= x3 + 53 - ( 8x3 + 12x2 + 6x + 1 ) - 28x3 - 33x2 + 15x
= -27x3 + 125 - 8x3 - 12x2 - 6x - 1 - 33x2 + 15x
= -33x3 - 45x2 + 9x + 124 ( có phụ thuộc vào biến )
B = ( 3x + 2 )3 - 18x( 3x + 2 ) + ( x - 1 )3 - 28x3 + 3x( x - 1 )
= 27x3 + 54x2 + 36x + 8 - 54x2 - 36x + x3 - 3x2 + 3x - 1 - 28x3 + 3x2 - 3x
= 7 ( đpcm )
C = ( 4x - 1 )( 16x2 + 4x + 1 ) - ( 4x + 1 )3 + 12( 4x + 1 )3 + 12( 4x + 1 ) - 15
= ( 4x )3 - 13 - [ ( 4x + 1 )3 - 12( 4x + 1 )3 - 12( 4x + 1 ) ] - 15
= 64x3 - 1 - ( 4x + 1 )[ ( 4x + 1 )2 - 12( 4x + 1 )2 - 12 ] - 15
= 64x3 - 16 - ( 4x + 1 )[ 16x2 + 8x + 1 - 12( 16x2 + 8x + 1 ) - 12 ]
= 64x3 - 16 - ( 4x + 1 )( 16x2 + 8x - 11 - 192x2 - 96x - 12 )
= 64x3 - 16 - ( 4x + 1 )( -176x2 - 88x - 23 )
= 64x3 - 16 - ( -704x3 - 528x2 - 180x - 23 )
= 64x3 - 16 + 704x3 + 528x2 + 180x + 23
= 768x3 + 528x2 + 180x + 7 ( có phụ thuộc vào biến )
a) \(\left(x+2\right)^2-9=0\)
\(=>\left(x+2\right)^2-3^2=0\\ =>\left(x+2-3\right).\left(x+2+3\right)=0\)
\(=>\left(x-1\right).\left(x+5\right)=0\)
\(=>\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}=>\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy x= 1 hoặc x= -5
b) \(x^2-2x+1=25\)
\(=>x^2-2.x.x+1^2=25\)
\(=>\left(x-1\right)^2-25=0\\ =>\left(x-1\right)^2-5^2=0\)
\(=>\left(x-1-5\right).\left(x-1+5\right)=0\)
\(=>\left(x-6\right).\left(x+4\right)=0=>\orbr{\begin{cases}x-6=0\\x+4=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
Vậy x= 6 hoặc x= -4
c) \(4x\left(x-1\right)-\left(2x+5\right)\left(2x-5\right)=1\)
\(=>4x\left(x-1\right)-\left[\left(2x\right)^2-5^2\right]=1\)
\(=>4x\left(x-1\right)-4x^2+25-1=0\)
\(=>4x\left(x-1\right)-4x^2+24=0\)
\(=>4x\left(x-1\right)-\left(4x^2-24\right)=0\\ =>4x\left(x-1\right)-4\left(x^2-6\right)=0\)
..................... tắc ròi -.-"
d) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+3\right)=15\)
\(=>x^3+27-x^3-3x=15\)
\(=>27-3x-15=0=>12-3x=0=>3\left(4-x\right)=0\)
Vì \(3>0=>4-x=0=>x=4\)
Vậy x= 4
e) \(3\left(x+2\right)^2+\left(2x+1\right)^2-7\left(x+3\right)\left(x-3\right)=28\)
\(=>3\left(x^2+2.x.2+2^2\right)+4x^2+4x+1-7\left(x^2-9\right)=28\)
\(=>3\left(x^2+4x+4\right)+4x^2+4x+1-7x^2+63=28\)
\(=>3x^2+12x+12+4x^2+4x+1-7x^2+63=28\)
\(=>16x+75=28=>16x=-47=>x=\frac{-47}{16}\)
Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt :>'-'
`#040911`
`a,`
`15 + 25 \div (2x - 1) = 20`
`\Rightarrow 25 \div (2x - 1) = 20 - 15`
`\Rightarrow 25 \div (2x - 1) = 5`
`\Rightarrow 2x - 1 = 25 \div 5`
`\Rightarrow 2x - 1 = 5`
`\Rightarrow 2x = 6`
`\Rightarrow x = 3`
Vây, `x = 3.`
`b,`
\(3^{x-1}+2\cdot3^x=21\)
`\Rightarrow 3^x \div 3 + 2. 3^x = 21`
`\Rightarrow 3^x . \frac{1}{3} + 2. 3^x = 21`
`\Rightarrow 3^x . (\frac{1}{3} + 2) = 21`
`\Rightarrow 3^x . \frac{7}{3} = 21`
`\Rightarrow 3^x = 21 \div \frac{7}{3}`
`\Rightarrow 3^x = 9`
`\Rightarrow 3^x = 3^2`
`\Rightarrow x = 2`
Vậy, `x = 2.`
`c,`
\(2^{x-3}+2^{x+1}=17\)
`\Rightarrow 2^x \div 2^3 + 2^x . 2 = 17`
`\Rightarrow 2^x . \frac{1}{8} + 2^x . 2 = 17`
`\Rightarrow 2^x . (\frac{1}{8} + 2) = 17`
`\Rightarrow 2^x . \frac{17}{8} = 17`
`\Rightarrow 2^x = 17 \div \frac{17}{8}`
`\Rightarrow 2^x = 8`
`\Rightarrow 2^x = 2^3`
`\Rightarrow x = 3`
Vậy, `x = 3`
`d,`
\(5^x-5^{x-1}=20\)
`\Rightarrow 5^x - 5^x \div 5 = 20`
`\Rightarrow 5^x - 5^x . \frac{1}{5} = 20`
`\Rightarrow 5^x . (1 - \frac{1}{5} = 20`
`\Rightarrow 5^x . \frac{4}{5} = 20`
`\Rightarrow 5^x = 20 \div \frac{4}{5}`
`\Rightarrow 5^x = 25`
`\Rightarrow 5^x = 5^2`
`\Rightarrow x = 2`
Vậy, `x = 2.`
\(a.25:\left(2x-1\right)=5\)
\(2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)
\(b.3^x:3+2.3^x=21\)\(\Leftrightarrow3^x.\dfrac{1}{3}+2.3^x=21\)
\(\Leftrightarrow3^x\left(\dfrac{1}{3}+2\right)=21\)
\(\Leftrightarrow3^x.\dfrac{7}{3}=21\)
\(\Leftrightarrow3^x=9\Leftrightarrow x=2\)
\(c.2^x:2^3+2^x.2=17\Leftrightarrow2^x.\dfrac{1}{8}+2^x.2=17\)
\(\Leftrightarrow2^x.\dfrac{17}{8}=17\Leftrightarrow2^x=8\Leftrightarrow x=3\)
\(d.5^x-5^x:5=20\Leftrightarrow5^x-5^x.\dfrac{1}{5}=20\)
\(\Leftrightarrow5^x\left(1-\dfrac{1}{5}\right)=20\Leftrightarrow5^x=20:\dfrac{4}{5}\Leftrightarrow5^x=25\Leftrightarrow x=2\)
`Answer:`
a. \(x+12=3\Leftrightarrow x=3-12\Leftrightarrow x=-9\)
b. \(2x-15=21\Leftrightarrow2x=21+15\Leftrightarrow2x=36\Leftrightarrow x=36:2\Leftrightarrow x=18\)
c. \(13-3x=4\Leftrightarrow-3x=4-13\Leftrightarrow-3x=-9\Leftrightarrow x=-9:-3\Leftrightarrow x=3\)
d. \(2\left(x-2\right)+4=12\Leftrightarrow2x-4+4=12\Leftrightarrow2x=12\Leftrightarrow x=12:2\Leftrightarrow x=6\)
e. \(15-3\left(x-2\right)=21\Leftrightarrow15-3x+6=21\Leftrightarrow-3x=21-15-6\Leftrightarrow-3x=0\Leftrightarrow x=0\)
g. \(25+4\left(3-x\right)=1\Leftrightarrow25+12-4x=1\Leftrightarrow37-4x=1\Leftrightarrow-4x=-36\Leftrightarrow x=9\)
h. \(3x+12=2x-4\Leftrightarrow3x-2x=-4-12\Leftrightarrow x=-16\)
i. \(14-3x=\left(-x\right)+4\Leftrightarrow-3x+x=4-14\Leftrightarrow-2x=10\Leftrightarrow x=5\)
k. \(2\left(x-2\right)+7=x-25\Leftrightarrow2x-4+7=x-25\Leftrightarrow2x-x=-25-3\Leftrightarrow x=-28\)
c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow3x^2+26x=0\)
\(\Leftrightarrow x\left(3x+26\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)
\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)
\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
17x + 3. ( -16x – 37) = 2x + 43 - 4x
<=>17x-48x-111=-2x+43
<=>-29x=154
<=> \(x=-\frac{154}{29}\)
-3. (2x + 5) -16 < -4. (3 – 2x)
\(\Leftrightarrow-6x-31< -12+8x.\)
\(\Leftrightarrow-14x< 19\Rightarrow x< -\frac{19}{14}\)