Tìm \(x\) sao cho :
\(\left|2x-4\right|=6\)
SK
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Tìm x:
\(3.\left(2x-6\right)-4.\left(1+2x\right)-2.\left(x-4\right)=4-3\left(1+2x\right)-5.\left(1-2x\right)\)
Ai nhanh mk tick cho . Mk cảm ơn bạn nhiều .
\(3\left(2x-6\right)-4\left(1+2x\right)-2\left(x-4\right)=4-3\left(1+2x\right)-5\left(1-2x\right).\)
\(\Leftrightarrow6x-18-4-8x-2x+8=4-3-6x-5+10x\)
\(\Leftrightarrow-4x-14=4x-4\)
\(\Leftrightarrow-4x-4x=-4+14\)
\(\Leftrightarrow-8x=10\)
\(\Leftrightarrow x=-\frac{5}{4}\)
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Tìm x :
\(3.\left(2x-6\right)-4.\left(1+2x\right)-2.\left(x-4\right)=4-3.\left(1+2x\right)-5\left(1-2x\right)\)
Ai nhanh mk tick cho mk cam on
3.2x - 3.6 - 4+4.2x - 2x-2.(-4) = 4 - 3+3.2x - 5-5.(-2x)
6x -18 -4 +8x -2x +8 = 4 -3 +6x -5 +10x
6x +8x -2x -18-4+8 = 4-3-5+6x+10x
12x-22 = -4+16x
12x-16x = -4+22
-4x = 18
x = 18: (-4)
x = -4,5
Mình không chắc là đúng đâu đấy, tại giải vội quá, nếu sai thì ming bạn thông cảm ^.^
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Tìm x
a)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)6
b)\(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
Nhanh nha kick cho :)
Ta có (x - 2)2 - (x - 3) (x + 3) = 6
=> (x - 2)2 - (x2 - 32) = 6
=> x2 - 4x + 22 - x2 + 32 = 6
=> x2 - x2 - 4x + 4 + 9 = 6
=> - 4x + 13 = 6
=> -4x = -7
=> x = \(\frac{-7}{-4}=\frac{7}{4}\)
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a) \(\Leftrightarrow\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow x^2-4x+4-x^2+9=6\)
\(\Leftrightarrow13-4x=6\)
\(\Leftrightarrow4x=7\)
\(\Leftrightarrow x=\frac{7}{4}\)
b) \(4(x−3)^2−(2x−1)(2x+1)=10\)
\(\Leftrightarrow4\left(x^2-6x+9\right)−4x^2+1=10\)
\(\Leftrightarrow4x^2-24x+36−4x^2+1=10\)
\(\Leftrightarrow37-24x=10\)
\(\Leftrightarrow24x=27\)
\(\Leftrightarrow x=\frac{9}{8}\)
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2. tìm x
a) \(\left(x-1\right)^3=8\)
b) \(7^{2x-6}=49\)
c) \(\left(2x-14\right)^7=128\)
d) \(x^4.x^5=5^3.5^6\)
e) \(\left[3.\left(x+2\right):7\right].4=120\)
a) \(\left(x-1\right)^3=8=2^3\)
\(x-1=2\)
\(x=2+1=3\)
b) \(7^{2x-6}=49=7^2\)
\(2x-6=2\)
\(2x=6+2=8\)
\(x=8:2=4\)
c) \(\left(2x-14\right)^7=128=2^7\)
\(2x-14=2\)
\(2x=14+2=16\)
\(x=16:2=8\)
d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)
\(x=5\)
e) \(3\cdot\left(x+2\right):7\cdot4=120\)
\(x+2=120:3\cdot7:4\)
\(x+2=70\)
\(x=70-2=68\)
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Lời giải:
a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$
$\Rightarrow x=3$
b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$
$\Rightarrow 2x=8$
$\Rightarrow x=4$
c. $(2x-14)^7=128=2^7$
$\Rightarrow 2x-14=2$
$\Rightarrow 2x=16$
$\Rightarrow x=18$
d.
$x^4.x^5=5^3.5^6$
$x^9=5^9$
$\Rightarrow x=5$
e.
$3(x+2):7=120:4=30$
$3(x+2)=30.7=210$
$x+2=210:3=70$
$x=70-2=68$
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tìm m để pt có nghiệm
\(6+x+2\sqrt{\left(4-x\right)\left(2x-2\right)}=m+4\left(\sqrt{4-x}+\sqrt{2x-2}\right)\)
a) Cho đa thức \(f\left(x\right)=\left(x-4\right)-3\left(x+1\right)\). Tìm x sao cho \(f\left(x\right)=4\)
b) Cho đa thức \(g\left(x\right)=m^2x^{10}+\left(3m+4\right)x^5+m^2x-10\). Tìm m biết rằng đa thức g (x) nhận x = -1 làm nghiệm.
KG
9 tháng 1 2024 lúc 21:30
Cho đa thức \(P\left(x\right)=\dfrac{5x^3}{4}+\dfrac{5x^2}{6}-\dfrac{21x}{4}+\dfrac{1}{6}\). Tìm số dư khi chia \(P\left(x\right)\) cho \(2x-5\).
Lời giải:
Theo định lý Bê-du về phép chia đa thức, số dư của $P(x)$ khi chia $2x-5$ là $P(\frac{5}{2})=\frac{5}{4}(\frac{5}{2})^3+\frac{5}{6}(\frac{5}{2})^2-\frac{21}{4}.\frac{5}{2}+\frac{1}{6}=\frac{377}{32}$
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BÀI 6 tìm x1,2xleft(x-5right)-left(3x+2x^2right)0 2,xleft(5-2xright)+2xleft(x-1right)133,2x^3left(2x-3right)-x^2left(4x^2-6x+2right)0 4,5xleft(x-1right)-left(x+2right)left(5x-7right)65,6x^2-left(2x-3right)left(3x+2right)1 6,2xleft(1-xright)+59-2x^2
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BÀI 6 tìm x
1,\(2x\left(x-5\right)-\left(3x+2x^2\right)=0\) 2,\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
3,\(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\) 4,\(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
5,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\) 6,\(2x\left(1-x\right)+5=9-2x^2\)
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
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`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`
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tìm x sao cho:
\(\frac{2+4+6+.....+2x}{1+3+5+......+\left(2x-1\right)}=\frac{16}{15}\)
nhanh thì có tick
Tìm x,y sao cho
\(\left(x^2+y+\frac{3}{4}\right)\left(y^2+x+\frac{3}{4}\right)=\left(2x+\frac{1}{2}\right)\left(2y+\frac{1}{2}\right)\)
\(x^2+y+\frac{3}{4}\ge x^2+\frac{1}{4}+y+\frac{1}{2}\ge2\sqrt{x^2\cdot\frac{1}{4}}+\left(y+\frac{1}{2}\right)\ge x+y+\frac{1}{2}\)
\(\Rightarrow VT\ge\left(x+y+\frac{1}{2}\right)^2=\left[\left(x+\frac{1}{4}\right)+\left(y+\frac{1}{4}\right)\right]^2\ge4\left(x+\frac{1}{4}\right)\left(y+\frac{1}{4}\right)\)
\(=\left(2x+\frac{1}{2}\right)\left(2y+\frac{1}{2}\right)\)
Dấu "=" xảy ra tại \(x=y=\frac{1}{2}\)
Vậy \(x=y=\frac{1}{2}\)
\(PT\Leftrightarrow x^2y^2+y^3+x^3+\frac{3}{4}\left(x^2+y^2\right)+xy+\frac{3}{4}\left(x+y\right)+\frac{9}{16}=4xy+x+y+\frac{1}{4}.\)
\(\Leftrightarrow x^2y^2+\left(x+y\right)^3-3xy\left(x+y\right)+\frac{3}{4}\left[\left(x+y\right)^2-2xy\right]+\frac{1}{4}\left(x+y\right)-3xy+\frac{5}{16}=0\)
Đặt \(x+y=a,xy=b\)
\(\Rightarrow b^2+a^3-3ab+\frac{3}{4}\left(a^2-2b\right)+\frac{a}{4}-3b+\frac{5}{16}=0\)
\(\Leftrightarrow16b^2+16a^3-48ab+12a^2-24b+4a-48b+5=0\)
\(\Leftrightarrow16b^2+16a^3-48ab+12a^2-72b+4a+5=0\)
Đến đây phân tích thành nhân tử hay sao ấy, chưa nghĩ ra :P