Giải các phương trình sau:
a \(x^2-11=0\)
b \(x^2-12x+52=0\)
c \(x^2-3x-28=0\)
d \(x^2-11x+38=0\)
e \(6x^2+71x+175=0\)
f \(x^2-\left(\sqrt{2}+\sqrt{8}\right)x+4=0\)
g\(\left(1+\sqrt{3}\right)x^2-\left(2\sqrt{3}+1\right)x+\sqrt{3}=0\)
giải phương trình :
a, \(2x^2-11x+21-3\sqrt[3]{4x-4}=0\)
b, \(\left(3x-2\right)\sqrt{x+1}-x^2-x-2=0\)
c, \(x+4-2\left(\dfrac{x+2}{x-1}\right)\sqrt{\dfrac{x-1}{x+2}}=0\)
c.
ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)
\(\Leftrightarrow x+4-2\sqrt[]{\left(\dfrac{x+2}{x-1}\right)^2\left(\dfrac{x-1}{x+2}\right)}=0\)
\(\Leftrightarrow x+4-2\sqrt[]{\dfrac{x+2}{x-1}}=0\)
\(\Leftrightarrow x+4=2\sqrt[]{\dfrac{x+2}{x-1}}\) (\(x\ge-4\))
\(\Leftrightarrow x^2+8x+16=\dfrac{4\left(x+2\right)}{x-1}\)
\(\Rightarrow x^3+7x^2+4x-24=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+4x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2+2\sqrt{3}\\x=-2-2\sqrt{3}\left(loại\right)\end{matrix}\right.\)
a.
\(\Leftrightarrow2x^2-11x+21=3\sqrt[3]{4\left(x-1\right)}\)
Do \(2x^2-11x+21=2\left(x-\dfrac{11}{4}\right)^2+\dfrac{47}{8}>0\Rightarrow3\sqrt[3]{4\left(x-1\right)}>0\Rightarrow x-1>0\)
Ta có:
\(VT=2x^2-11x+21-3\sqrt[3]{4x-4}=2\left(x^2-6x+9\right)+x+3-3\sqrt[3]{4\left(x-1\right)}\)
\(=2\left(x-3\right)^2+x+3-3\sqrt[3]{4\left(x-1\right)}\)
\(\Rightarrow VT\ge x+3-3\sqrt[3]{4\left(x-1\right)}=\left(x-1\right)+2+2-3\sqrt[3]{4\left(x-1\right)}\)
\(\Rightarrow VT\ge3\sqrt[3]{\left(x-1\right).2.2}-3\sqrt[3]{4\left(x-1\right)}=0\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\x-1=2\\\end{matrix}\right.\) \(\Leftrightarrow x=3\)
Vậy pt có nghiệm duy nhất \(x=3\)
b.
ĐKXD: \(x\ge-1\)
Phương trình: \(2\left(x+1\right)-\left(3x-2\right)\sqrt[]{x+1}+x^2-x=0\)
Đặt \(\sqrt[]{x+1}=t\ge0\)
\(\Rightarrow2t^2-\left(3x-2\right)t+x^2-x=0\)
\(\Delta=\left(3x-2\right)^2-8\left(x^2-x\right)=\left(x-2\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x-2+x-2}{4}=x-1\\t=\dfrac{3x-2-x+2}{4}=\dfrac{x}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x+1}=x-1\left(x\ge1\right)\\\sqrt[]{x+1}=\dfrac{x}{2}\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=x^2-2x+1\left(x\ge1\right)\\x+1=\dfrac{x^2}{4}\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2+2\sqrt[]{2}\end{matrix}\right.\)
Giải phương trình sau:
a) \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
b) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
c) \(2x-x^2+\sqrt{6x^2-12x+7}=0\)
d) \(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
Dùng công thức nghiệm,công thức nghiệm thu gọn giải các phương trình sau:
a.\(x^2-4x-21=0\)
b.\(4x^2+28x+49=0\)
c.\(6y^2-5\sqrt{2}y+2=0\)
d.\(y^2-\left(1+\sqrt{3}\right)y+\sqrt{3}=0\)
e.\(x^2+3x-10=0\)
a: =>(x-7)(x+3)=0
hay \(x\in\left\{7;-3\right\}\)
b: =>2x+7=0
hay x=-7/2
c: \(\Delta=50-4\cdot6\cdot2=50-48=2\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{5\sqrt{2}-\sqrt{2}}{12}=\dfrac{\sqrt{2}}{3}\\x_2=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
Giải phương trình :
a.\(x^2+5x^2-3=0\)
b.\(x^2-\left(2\sqrt{3}-1\right)x+4\sqrt{3}-6=0\)
c.\(x^2-6x+9=0\)
d.\(x^2-4\sqrt{3}x-4=0\)
c: \(\Leftrightarrow x-3=0\)
hay x=3
Giải các pt sau:
a)\(\left|3x+1\right|=\left|x+1\right|\)
b)\(\left|x^2-3\right|=\left|x-\sqrt{3}\right|\)
c)\(\sqrt{9x^2-12x+4}=\sqrt{x^2}\)
d)\(\sqrt{x^2+4x+4}=\sqrt{4x^2-12x+9}\)
e) \(\left|x^2-1\right|+\left|x+1\right|=0\)
f)\(\sqrt{x^2-8x+16}+\left|x+2\right|=0\)
g) \(\sqrt{1-x^2}+\sqrt{x+1}=0\)
h) \(\sqrt{x^2-4}+\sqrt{x^2+4x+4}=0\)
Mọi người giúp em gấp với!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
a) \(\left|3x+1\right|=\left|x+1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=x+1\\3x+1=-x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
c) \(\sqrt{9x^2-12x+4}=\sqrt{x^2}\)
\(\Leftrightarrow\sqrt{\left(3x-2\right)^2}=\sqrt{x^2}\)
\(\Leftrightarrow\left|3x-2\right|=\left|x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=x\\3x-2=-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
d) \(\sqrt{x^2+4x+4}=\sqrt{4x^2-12x+9}\)
\(\Leftrightarrow\sqrt{\left(x+2\right)^2}=\sqrt{\left(2x-3\right)^2}\)
\(\Leftrightarrow\left|x+2\right|=\left|2x-3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-3\\x+2=-2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\left|x^2-1\right|+\left|x+1\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-1\)
f) \(\sqrt{x^2-8x+16}+\left|x+2\right|=0\)
\(\Leftrightarrow\sqrt{\left(x-4\right)^2}+\left|x+2\right|=0\)
\(\Leftrightarrow\left|x-4\right|+\left|x+2\right|=0\)
⇒ vô nghiệm
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
Giải các pt sau:
a) \(\sqrt{x+8}+\frac{9x}{\sqrt{x+8}}-6\sqrt{x}=0\)
b) \(x^4-2x^3+\sqrt{2x^3+x^2+2}-2=0\)
c) \(3x\sqrt[3]{x+7}\left(x+\sqrt[3]{x+7}\right)=7x^3+12x^2+5x-6\)
d) \(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
e) \(16x^2+19x+7+4\sqrt{-3x^2+5x+2}=\left(8x+2\right)\left(\sqrt{2-x}+2\sqrt{3x+1}\right)\)
f) \(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+8-\left(x+26\right)\sqrt{x-1}\)
g) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
Dùng công thức nghiệm,giải các phương trình sau:
a. \(x^2+3x+4=0\)
b. \(4x^2-4x+1=0\)
c. \(x^2-5x-6=0\)
d. \(3x^2+12x-2=0\)
e. \(x^2+2\sqrt{5}x-1=0\)
f. \(2x^2-4\sqrt{2}x+2=0\)
Giair phương trình sau:
a,\(2x^3+5x^2-3x=0\) b,\(2x^3+6x^2=x^2+3x\)
c,\(x^2+\left(x+2\right)\left(11x-7\right)=4\) d,\(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)
e, \(x^3+1=x\left(x+1\right)\) f,\(x^3+x^2+x+1=0\)
g,\(x^3-3x^2+3x-1=0\) h,\(x^3-7x+6=0\)
i,\(x^6-x^2=0\) j,\(x^3-12=13x\)
k,\(-x^5+4x^4=-12x^3\) l, \(x^3=4x\)
a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
Trong đó có nhiều phương trình kiến thức cơ bản mà nhỉ? Ít nâng cao, bạn lọc ra câu nào k làm đc thôi chứ!