Giải phương trình:
\(\sqrt[3]{3x^2-2x+2017}-\sqrt[3]{3x^2-8x+2018}-\sqrt[3]{6x-2019}=\sqrt[3]{2018}\)
1.Giải hệ phương trình
\(\hept{\begin{cases}y^3+\sqrt{8x^4-2y}=2\left(2x^4+3\right)\\\sqrt{2x^2+x+y}+2\sqrt{x+2y}=\sqrt{9x-2x^2+17y}\end{cases}}\)
2.Cho P(x) là đa thức bậc 3 có hệ số bậc cao nhất là 1 và thảo mãn:
P(2016)=2017;P(2017)=2018.Tính:-3P(2018)+P(2019)
3.Cho x,y,z\(\ge1\)thỏa mãn:\(3x^2+4y^2+5Z^2=32\)
Tìm min:x+y+z
Giải phương trình:
a) \(2\sqrt{x^2-4}-3=6\sqrt{x-2}-\sqrt{x+2}\)
b) \(\frac{\sqrt{x-2016}-1}{x-2016}+\frac{\sqrt{y-2017}-1}{y-2017}+\frac{\sqrt{z-2018}-1}{z-2018}=\frac{3}{4}\)
c) \(\sqrt{3+\sqrt{3+x}}=x\)
d) \(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
e) \(\sqrt{x^2+3x+5}+\sqrt{x^2-2x+5}=5\sqrt{x}\)
f) \(\sqrt{x^2+3x}+2\sqrt{x+2}=2x+\sqrt{x+\frac{6}{x}+5}\)
a/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow2\sqrt{\left(x-2\right)\left(x+2\right)}-6\sqrt{x-2}+\sqrt{x+2}-3=0\)
\(\Leftrightarrow2\sqrt{x-2}\left(\sqrt{x+2}-3\right)+\sqrt{x+2}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x-2}+1\right)\left(\sqrt{x+2}-3\right)=0\)
\(\Leftrightarrow\sqrt{x+2}-3=0\Rightarrow x=11\)
b/ ĐKXĐ: ....
Đặt \(\left\{{}\begin{matrix}\sqrt{x-2016}=a>0\\\sqrt{y-2017}=b>0\\\sqrt{z-2018}=a>0\end{matrix}\right.\)
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{a-1}{a^2}+\frac{1}{4}-\frac{b-1}{b^2}+\frac{1}{4}-\frac{c-1}{c^2}=0\)
\(\Leftrightarrow\frac{\left(a-2\right)^2}{a^2}+\frac{\left(b-2\right)^2}{b^2}+\frac{\left(c-2\right)^2}{c^2}=0\)
\(\Leftrightarrow a=b=c=2\Rightarrow\left\{{}\begin{matrix}x=2020\\y=2021\\z=2022\end{matrix}\right.\)
a/ ĐK: \(x\ge0\)
\(\Leftrightarrow\sqrt{3+x}=x^2-3\)
Đặt \(\sqrt{3+x}=a>0\Rightarrow3=a^2-x\) pt trở thành:
\(a=x^2-\left(a^2-x\right)\)
\(\Leftrightarrow x^2-a^2+x-a=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a+1\right)=0\)
\(\Leftrightarrow x=a\) (do \(x\ge0;a>0\))
\(\Leftrightarrow\sqrt{3+x}=x\Leftrightarrow x^2-x-3=0\)
d/ ĐKXĐ: ...
\(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
\(\Leftrightarrow\sqrt{2x-3}-1+x^2+1-\sqrt{6x^2+1}\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^4+2x^2+1-6x^2-1}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)\left(x-2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{2}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}\right)=0\)
\(\Leftrightarrow x=2\) (phần trong ngoặc luôn dương với mọi \(x\ge\frac{3}{2}\))
e/ ĐKXĐ: \(x\ge0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+3x+5}=a>0\\\sqrt{x^2-2x+5}=b>0\\\sqrt{x}=c\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=5c^2\)
Ta được hệ: \(\left\{{}\begin{matrix}a^2-b^2=5c^2\\a+b=5c\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=5c^2\\a+b=5c\end{matrix}\right.\)
\(\Rightarrow5c\left(a-b\right)=5c^2\)
\(\Leftrightarrow\left[{}\begin{matrix}c=0\\a-b=c\end{matrix}\right.\)
f/ ĐKXĐ: \(x>0\)
\(\Leftrightarrow\sqrt{x\left(x+3\right)}+2\sqrt{x+2}=2x+\sqrt{\frac{\left(x+2\right)\left(x+3\right)}{x}}\)
\(\Leftrightarrow\sqrt{\frac{\left(x+2\right)\left(x+3\right)}{x}}-2\sqrt{x+2}+2x-2\sqrt{x\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{\frac{x+2}{x}}\left(\sqrt{x+3}-2\sqrt{x}\right)-2\sqrt{x}\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left(\sqrt{\frac{x+2}{x}}-2\sqrt{x}\right)\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x+3}{x}=4x\\x+3=4x\end{matrix}\right.\)
1. Giải phương trình sau:
\(x^3-3x^2+2\sqrt{\left(x+2\right)^3}-6x=0\)
2. Cho các số thực x,y thỏa mã điều kiện:
\(\sqrt{x^2+11}+\sqrt{x^2-2018}+x^2=\sqrt{y^2+11}+\sqrt{y^2-2018}+y^2\)
Tính giá trị biểu thức: \(M=x^{11}-y^{2018}\)
3. Cho tam giác ABC vuông tại A trên cạnh BC lấy điểm D bất kỳ. Gọi E và F lần lượt là hình chiếu của D trên cạnh AB và AC.
a) CM: DB.DC=EA.EB+FA.FC
b) Trên cạnh BC lấy điểm M sao cho ^BAD=^CAM
CMR: \(\dfrac{DB}{DC}.\dfrac{MB}{MC}=\dfrac{AB^2}{AC^2}\)
1.
đk: \(x\ge2\)
Đặt y = \(\sqrt{x+2}\) ta biến pt về dạng pt thuần nhất bậc 3 đối vs x và y:
ta có : \(x^3-3x^2+2y^3-6x=0\)
\(\Leftrightarrow x^3-3xy^2+2y^3=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\\x=-2y\end{matrix}\right.\)
ta sẽ có nghiệm : \(x=2;x=2-2\sqrt{3}\)
\(1.đk:\left(x+2\right)^3\ge0\Leftrightarrow x\ge-2\)
\(pt\Leftrightarrow x^3-3x\left(x+2\right)+2\sqrt{\left(x+2\right)^3}=0\)
\(\Leftrightarrow x^3-x\left(x+2\right)+2\sqrt{\left(x+3\right)^2}-2x\left(x+2\right)=0\)
\(\Leftrightarrow x\left[x^2-\left(x+2\right)\right]+2\left(x+2\right)\left(\sqrt{x+2}-x\right)=0\)
\(\Leftrightarrow x\left[\left(x-\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right)\right]+2\left(x+2\right)\left(\sqrt{x+2}-x\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+2}-x\right)\left[-x\left(\sqrt{x+2}+x\right)+2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(\sqrt{x+2}-x\right)^2\left(2\sqrt{x+2}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}=x\left(2\right)\\2\sqrt{x+2}=-x\left(3\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=x+2\end{matrix}\right.\)\(\Leftrightarrow x=2\left(tm\right)\)
\(\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}-x\ge0\Leftrightarrow x\le0\\x^2=4\left(x+2\right)\end{matrix}\right.\)\(\Leftrightarrow x=2-2\sqrt{3}\left(tm\right)\)
\(2.đk:x^2;y^2\ge2018\Leftrightarrow\left[{}\begin{matrix}x;y\le-\sqrt{2018}\\x;y\ge\sqrt{2018}\end{matrix}\right.\)
\(pt\Leftrightarrow\sqrt{x^2+11}-\sqrt{y^2+11}+\sqrt{x^2-2018}-\sqrt{y^2-2018}+x^2-y^2=0\)
\(\Leftrightarrow\left(x+y\right)\left(x-y\right)+\dfrac{x^2+11-y^2-11}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\dfrac{x^2-2018-y^2+2018}{\sqrt{x^2-2018}+\sqrt{y^2-2018}}=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left[1+\dfrac{1}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\dfrac{1}{\sqrt{x^2-2018}+\sqrt{y^2+2018}}>0\right]=0\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(x=y\Rightarrow M=x^{11}-x^{2018}\)
\(x=-y\Rightarrow M=-y^{11}-y^{2018}=:vvv\) (đến đây chịu)
giải phương trình sau:
a) \(4x^2+\left(8x-4\right).\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
b) \(8x^3-36x^2+\left(1-3x\right)\sqrt{3x-2}-3\sqrt{3x-2}+63x-32=0\)
c) \(2\sqrt[3]{3x-2}-3\sqrt{6-5x}+16=0\)
d) \(\sqrt[3]{x+6}-2\sqrt{x-1}=4-x^2\)
giải phương trình:
\(\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}=8x-4x^2-1\)
Giải các phương trình sau
\(1)\sqrt{3x+1}+\sqrt{5x+4}=3x^2-x+3\)
\(2)\left(4x-1\right)\sqrt[3]{2-8x^3}=2x\)
1.
ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(\Leftrightarrow3x^2-3x+\left(x+1-\sqrt{3x+1}\right)+\left(x+2-\sqrt{5x+4}\right)=0\)
\(\Leftrightarrow3\left(x^2-x\right)+\dfrac{x^2-x}{x+1+\sqrt{3x+1}}+\dfrac{x^2-x}{x+2+\sqrt{5x+4}}=0\)
\(\Leftrightarrow\left(x^2-x\right)\left(3+\dfrac{1}{x+1+\sqrt{3x+1}}+\dfrac{1}{x+2+\sqrt{5x+4}}\right)=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow...\)
2.
Đặt \(\left\{{}\begin{matrix}2x=a\\\sqrt[3]{2-8x^3}=b\end{matrix}\right.\)
Ta được hệ:
\(\left\{{}\begin{matrix}\left(2a-1\right)b=a\\a^3+b^3=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2ab\\\left(a+b\right)^3-3ab\left(a+b\right)=2\end{matrix}\right.\)
\(\Rightarrow8\left(ab\right)^3-6\left(ab\right)^2=2\)
\(\Leftrightarrow\left(ab-1\right)\left[4\left(ab\right)^2+ab+1\right]=0\)
\(\Leftrightarrow ab=1\Rightarrow a+b=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2\\ab=1\end{matrix}\right.\) \(\Leftrightarrow a=b=1\)
\(\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)
Cho \(x=1-\sqrt[3]{2}+\sqrt[3]{4}\)Tính \(B=x^{2019}-3x^{2018}+9x^{2017}-9x^{2016}+2019\)
\(x=1-\sqrt[2]{2}+\sqrt[2]{4}\)
\(\Leftrightarrow x\left(\sqrt[3]{2}+1\right)=\left(1-\sqrt[2]{2}+\sqrt[2]{4}\right)\left(\sqrt[3]{2}+1\right)=3\)
\(\Leftrightarrow\sqrt[3]{2}x=3-x\)
\(\Leftrightarrow2x^3=27-27x+9x^2-x^3\)
\(\Leftrightarrow x^3-3x^2+9x-9=0\)
Giờ tự rap xô vô nhe
giải các phương trình sau:
\(1,\sqrt{18x}-6\sqrt{\dfrac{2x}{9}}=3-\sqrt{\dfrac{x}{2}}\)
\(2,\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\sqrt{27x}=-4\)
3, \(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
\(4,\sqrt{16x+16}-\sqrt{9x+9}=1\)
\(5,\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
\(6,\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=\dfrac{-2}{3}\)
2: ĐKXĐ: x>=0
\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)
=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)
=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)
=>\(-2\sqrt{3x}=-4\)
=>\(\sqrt{3x}=2\)
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
3:
ĐKXĐ: x>=0
\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)
=>\(13\sqrt{2x}=20+3\sqrt{2}\)
=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)
=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)
=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)
4: ĐKXĐ: x>=-1
\(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
5: ĐKXĐ: x<=1/3
\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)
=>\(5\sqrt{1-3x}=10\)
=>\(\sqrt{1-3x}=2\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1(nhận)
6: ĐKXĐ: x>=3
\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)
=>x-3=16
=>x=19(nhận)
Giải các phương trình dưới đây
1, \(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)
2,\(\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}=2-x^2+2x\)
3, \(\sqrt{6y-y^2-5}-\sqrt{x^2-6x+10}=1\) (x=3 ; y=3)